网络流24题
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①飞行员配对方案【二分匹配】
洛谷P2756
二分图匹配 + 输出方案
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 305,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,N,M;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int main(){ memset(h,-1,sizeof(h)); M = RD(); N = RD(); S = 0; T = N + M + 1; int a,b; while (true){ a = RD(); b = RD(); if (a == -1 && b == -1) break; build(a,b,1); } REP(i,M) build(S,i,1); REP(i,N) build(M + i,T,1); int ans = maxflow(); if (!ans) printf("No Solution!\n"); else { printf("%d\n",ans); REP(i,M) Redge(i) if (!(k & 1) && !ed[k].f){ printf("%d %d\n",i,ed[k].to); break; } } return 0;}②太空飞行计划问题【最大权闭合子图】
经典的最大权闭合子图问题【记得我以前写过这个博客】
选一个正权物品就要附属上一些负权物品
我们将原图的边变为INF
之后由S引向正权的点，流量为正权权值
引负权的点向T，流量也为权值【相反数，也就是正的】
这样我们假设我们已经获得了所有正权，我们的损失就是最小割
对于一个正权物品，我们要么丢弃，要么选择支付其附属品
最后我们再跑一次dfs，S能到达的物品就是被选中的
【MMP我还没跑最大流就开始跑残量网络。。。查了半天】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 505,maxm = 100005,INF = 0x3f3f3f3f;int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,N,M,sum = 0;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(d); queue<int> q; d[S] = 1; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); if (u == T) return 1; Redge(u) if (ed[k].f && !d[to = ed[k].to]){ d[to] = d[u] + 1; q.push(to); } } return 0;}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}void DFS(int u){ int to; vis[u] = true; Redge(u) if (ed[k].f > 0 && !vis[to = ed[k].to]) DFS(to);}int main(){ memset(h,-1,sizeof(h)); int x,flow; scanf("%d%d",&M,&N); S = 0; T = N + M +1; for(int i=1;i<=M;++i) { scanf("%d",&x),build(S,i,x),sum+=x; char ch=getchar(); while(ch!='\n'&&ch!='\r'&&ch!=EOF) { x=0; while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); if(x) build(i,M+x,INF); if(ch!='\n'&&ch!='\r'&&ch!=EOF) ch=getchar(); } } for(int i=1;i<=N;++i) scanf("%d",&x),build(i+M,T,x); flow = maxflow(); DFS(S); REP(i,M) if (vis[i]) printf("%d ",i); putchar('\n'); REP(i,N) if (vis[M + i]) printf("%d ",i); putchar('\n'); printf("%d",sum - flow); return 0;}③最小路径覆盖问题
有两种类型：
①在DAG图上，选择多条路径【可以只含一个点】，路径经过的点不能有交集，求最小覆盖所有点的路径数
②在DAG图上，选择多条路径【可以只含一个点】，路径经过的点能有交集，求最小覆盖所有点的路径数
对于问题①，我们先把所有点看做一条路径，然后沿边外扩展，一开始答案是点数，最多能扩展多少答案就减少多少条路径
对于问题②，允许跨过重复的点，我们求一个传递闭包，实际求最大流时就可以跨过重复的点了
问题①洛谷P2764
问题②BZOJ1143
这里粘问题①代码：
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 505,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,N,M;int pre[maxn],nxt[maxn];struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int main(){ memset(h,-1,sizeof(h)); N = RD(); M = RD(); S = 0; T = 2 * N + 1; int a,b; REP(i,N) build(S,i,1),build(i + N,T,1); while (M--){ a = RD(); b = RD(); build(a,b + N,1); } int ans = maxflow(); REP(i,N) Redge(i) if (!(k & 1) && !ed[k].f) nxt[i] = ed[k].to - N,pre[ed[k].to - N] = true; REP(i,N) if (!pre[i]){ int u = i; while (u) printf("%d ",u),u = nxt[u]; printf("\n"); } printf("%d",N - ans); return 0;}④ 魔术球问题 【最小路径覆盖】
洛谷P2765
其实也是最小路径覆盖，但求的是在最小路径覆盖数一定情况下所取最多点数
我们就依次加入点以及对应边，在残量网络上跑，其实效率挺高
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 4005,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,s = 0;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 1; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}void init(){ build(S,2 * s - 1,1); build (2 * s,T,1); for (int i = 1; i < s; i++){ int v = (int)sqrt(i + s); if (v * v == i + s) build(2 * i - 1,2 * s,1); }}int pre[maxn],nxt[maxn];void getans(){ for (int i = 1; i < s; i++){ int u = 2 * i - 1; Redge(u) if (!(k & 1) && !ed[k].f){ nxt[i] = ed[k].to >> 1; pre[ed[k].to >> 1] = true; break; } } for (int i = 1; i < s; i++){ if (pre[i]) continue; int u = i; while (u) printf("%d ",u),u = nxt[u]; printf("\n"); }}int main(){ memset(h,-1,sizeof(h)); int n = RD(),flow = 0; S = 0; T = 3501; while (true){ s++; init(); flow += maxflow(); if (s - flow > n) break; } printf("%d\n",s - 1); getans(); return 0;}今天先到这
⑤圆桌问题
题意：M团人坐N个大小不同的桌，同一团的人不能坐同一桌，求是否有解及一个方案
QAQ找不到OJ提交，就先放着代码吧
建图：
①S->每个团，容量为人数
②圆桌->T，容量为桌子容量
③每个团->每个圆桌，容量为1
跑一遍最大流，看看S的出边是否全都满载
对于每个团，流向的圆桌就是方案
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 305,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,N,M;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int main(){ memset(h,-1,sizeof(h)); int M = RD(),N = RD(),tot = 0,x; S = 0; T = N + M + 1; REP(i,M) build(S,i,x = RD()),tot += x; REP(i,N) build(i + M,T,RD()); REP(i,M) REP(j,N) build(i,M + j,1); int ans = maxflow(); if (ans != tot) {printf("0"); return 0;} printf("1\n"); REP(i,M){ Redge(i) if (!(k & 1) && !ed[k].f) printf("%d ",ed[k].to - M); putchar('\n'); } return 0;}⑥最长递增子序列问题
洛谷2766
求出序列中每个数最多选一次，最多能组成多少个最长递增子序列
我们先DP算出最长长度K，然后建图：
S->f为1的点，容量1
f为K的点->T，容量1
所有f[i] == f[j] - 1，i->j，容量1
跑一遍最大流即可
如果首尾能使用多次，将其对应和源汇点的边改为INF即可
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int maxn=1005,INF=2000000000;inline int read(){ int out=0,flag=1;char c=getchar(); while(c<48||c>57) {if(c=='-') flag=-1;c=getchar();} while(c>=48&&c<=57) {out=out*10+c-48;c=getchar();} return out*flag;}int A[maxn],dp[maxn],N,K=0;int head[maxn],nedge=0;struct EDGE{ int to,f,c,next;}edge[maxn*maxn];inline void build(int a,int b,int w){ edge[nedge]=(EDGE){b,0,w,head[a]}; head[a]=nedge++; edge[nedge]=(EDGE){a,0,0,head[b]}; head[b]=nedge++;}bool vis[maxn];int d[maxn],S,T,cur[maxn],lans;bool bfs(){ fill(vis,vis+maxn,false); queue<int> q; q.push(S); vis[S]=true; d[S]=0; int u,to; while(!q.empty()){ u=q.front(); q.pop(); for(int k=head[u];k!=-1;k=edge[k].next) if(!vis[to=edge[k].to]&&edge[k].f<edge[k].c){ d[to]=d[u]+1; vis[to]=true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if(u==T||!minf) return minf; int f,flow=0,to; if(cur[u]==-2) cur[u]=head[u]; for(int& k=cur[u];k!=-1;k=edge[k].next) if(d[to=edge[k].to]==d[u]+1&&(f=dfs(to,min(minf,edge[k].c-edge[k].f)))){ edge[k].f+=f; edge[k^1].f-=f; flow+=f; minf-=f; if(!minf) break; } return flow;}int maxflow(){ int flow=0; while(bfs()){ fill(cur,cur+maxn,-2); flow+=dfs(S,INF); } return flow;}void solve1(){ S=0; T=2*N+1; for(int i=1;i<=N;i++){ build(i,i+N,1); if(dp[i]==1) build(S,i,1); if(dp[i]==K) build(i+N,T,1); for(int j=1;j<i;j++) if(A[j]<=A[i]&&dp[j]==dp[i]-1) build(j+N,i,1); } lans=maxflow(); cout<<lans<<endl;}void solve2(){ for(int k=head[S];k!=-1;k=edge[k].next) if(edge[k].to==1){edge[k].c=INF;break;} for(int k=head[1];k!=-1;k=edge[k].next) if(edge[k].to==N+1){edge[k].c=INF;break;} for(int k=head[N];k!=-1;k=edge[k].next) if(edge[k].to==N+N){edge[k].c=INF;break;} for(int k=head[N+N];k!=-1;k=edge[k].next) if(edge[k].to==T){edge[k].c=INF;break;} cout<<lans+maxflow()<<endl;}int main(){ fill(head,head+maxn,-1); N=read(); for(int i=1;i<=N;i++) A[i]=read(); for(int i=1;i<=N;i++){ dp[i]=1; for(int j=1;j<i;j++) if(A[j]<=A[i]) dp[i]=max(dp[i],dp[j]+1); } for(int i=1;i<=N;i++) if(dp[i]>K) K=dp[i]; cout<<K<<endl; solve1(); solve2(); return 0;}⑦试题库问题
洛谷P2763
每种类型选择一定数量，每种类型都有对应的一些题可以选择
实际就是多匹配的问题
建图：
①S->类型，容量为对应需要的数量
②题目->T，容量为1【每个题只能用一次】
③类型->对应题目，容量1
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 1305,maxm = 1000005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int main(){ memset(h,-1,sizeof(h)); int K = RD(),N = RD(),M = 0,x,p; S = 0; T = K + N + 1; REP(i,K) build(S,i,x = RD()),M += x; REP(i,N){ p = RD(); build(K + i,T,1); while (p--) build(RD(),K + i,1); } int ans = maxflow(); if (ans != M) {printf("No Solution!"); return 0;} REP(i,K){ printf("%d:",i); Redge(i) if (!(k & 1) && !ed[k].f){ printf(" %d",ed[k].to - K); } putchar('\n'); } return 0;}⑧机器人路径规划问题
听说很难
⑨方格取数问题【最大点独立集】
BZOJ1475 洛谷2774
方格取数，所取数不能相邻
我们对矩阵黑白染色就会发现同种颜色互不干涉，相邻的黑白只能取一
这就是最大点独立集呐
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 1005,maxm = 200005,INF = 0x7fffffff;inline LL RD(){ LL out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T;struct EDGE{int to,nxt; LL f;}ed[maxm];inline void build(int u,int v,LL w){ ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++; ed[ne] = (EDGE){u,h[v],0}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}LL dfs(int u,LL minf){ if (u == T || !minf) return minf; LL flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}LL maxflow(){ LL flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int main(){ memset(h,-1,sizeof(h)); int n = RD(),N = n * n; S = 0; T = N + 1; LL x,sum = 0; REP(i,n) REP(j,n){ int u = n * (i - 1) + j; if ((i & 1) ^ (j & 1)){ build(S,u,x = RD()); if (i > 1) build(u,u - n,INF); if (i < n) build(u,u + n,INF); if (j > 1) build(u,u - 1,INF); if (j < n) build(u,u + 1,INF); } else build(u,T,x = RD()); sum += x; } cout<<sum - maxflow()<<endl; return 0;}⑩餐巾计划问题【线性规划与网络流】
洛谷P1251
问题见原题。【大概就是每天要用一定量的餐巾，要么直接购买，要么通过之前用的送到快洗慢洗店洗得，求最小费用】
想清楚的话建图还是很容易想到的：
拆点，一分为三，记为xi,yi,zi，分别表示购买餐巾、使用餐巾、送洗餐巾
建边：
①S->xi，容量INF，费用0【商店的餐巾】
②yi->T，容量Ri，费用0【使用的餐巾】
③xi->yi，容量INF，费用pi【由商店到餐厅，需要消费】
④xi->zi，容量Ri，费用0【每天会产生Ri个使用过的餐巾】
⑤zi->xj，有两条，容量INF，费用f或s，分别指向i + m和i + n【如果送洗，完成后立刻拿回来】
⑥yi->y(i+1)，容量INF，费用0【放着不用的纸巾】
跑一遍费用流，就可以得到答案【很形象的其实，有没有】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 6005,maxm = 100005;LL INF = 10000000000000000ll;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,pa[maxn],minf[maxn],R[maxn];LL d[maxn];bool inq[maxn];struct EDGE{int from,to,nxt,f; LL w;}ed[maxm];inline void build(int u,int v,int f,LL w){ ed[ne] = (EDGE){u,v,h[u],f,w}; h[u] = ne++; ed[ne] = (EDGE){v,u,h[v],0,-w}; h[v] = ne++;}LL mincost(){ queue<int> q; LL u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; pa[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[pa[u]].f -= minf[T]; ed[pa[u]^1].f += minf[T]; u = ed[pa[u]].from; } } return cost;}int main(){ memset(h,-1,sizeof(h)); int N = RD(); S = 0; T = 3 * N + 1; REP(i,N) build(S,i,INF,0),build(i + N,T,R[i] = RD(),0); int p = RD(),m = RD(),f = RD(),n = RD(),s = RD(); REP(i,N){ build(i,i + N,INF,p); build(i,i + 2 * N,R[i],0); if (i + m <= N) build(i + 2 * N,i + m + N,INF,f); if (i + n <= N) build(i + 2 * N,i + n + N,INF,s); if (i < N) build(i + N,i + 1 + N,INF,0); } printf("%lld\n",mincost()); return 0;}⑪航空路线问题 【最大费用最大流】
洛谷P2770
寻找DAG图中从一个点出发到终点的两条不相交路径，使得经过尽量多的点
我们拆点使得每个点流量1费用为1，起点终点流量为2，其他边流量1费用0，跑一遍最大费用最大流【就是每次找DAG最长路】
【坑点，ST直接相连，流量要为2】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<map>#include<string>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 305,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}map<string,int> H;string name[maxn];int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn],id = 0;bool inq[maxn],vis[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int maxcost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = S; i <= T; i++) d[i] = -1,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] < d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == -1) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u != S){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int code(const string& s){ if (!H.count(s)) H[s] = ++id,name[id] = s; return H[s];}int main(){ memset(h,-1,sizeof(h)); int N = RD(),V = RD(),a,b; S = 1; T = 2 * N; string s; REP(i,N) cin>>s,code(s); build(1,N + 1,2,1); build(N,N + N,2,1); for (int i = 2; i < N; i++) build(i,i + N,1,1); while (V--){ cin>>s; a = code(s); cin>>s; b = code(s); if (a > b) swap(a,b); if (a == 1 && b == N) build(a + N,b,2,0); else build(a + N,b,1,0); } int ans = maxcost() - 2; if (ed[h[S]].f) {printf("No Solution!\n"); return 0;} printf("%d\n",ans); int u = 1 + N,to; cout<<name[1]<<endl; while (u != N + N){ Redge(u) if (!(k & 1) && !ed[k].f && !vis[to = ed[k].to]){ cout<<name[to]<<endl; vis[to] = true; u = to + N; break; } } u = N; while (u != S){ Redge(u) if ((k & 1) && !ed[k ^ 1].f && !vis[to = (ed[k].to - N)]){ cout<<name[to]<<endl; vis[to] = true; u = to; break; } } return 0;}⑫软件补丁问题【最小代价转移】
洛谷P2761
这。。。网络流？
状压最短路可以写，1表示有该BUG，0表示没有改BUG，跑一遍SPFA
对于每个点，检查每一个补丁，看看是否满足B1含有而B2 不含有
然后去掉F1(u^F&u)并上F2
用到一些位运算技巧
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 105,maxm = 1 << 20,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int w[maxn],B1[maxn],B2[maxn],F1[maxn],F2[maxn],n,m,d[maxm];bool inq[maxm];queue<int> q;void SPFA(){ fill(d,d + maxm,INF); q.push((1 << n) - 1); inq[(1 << n) - 1] = true; d[(1 << n) - 1] = 0; int u,to; while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; REP(i,m) if ((u & B1[i]) == B1[i] && !(u & B2[i])){ to = (u ^ F1[i] & u) | F2[i]; if (d[u] + w[i] < d[to]){ d[to] = d[u] + w[i]; if (!inq[to]) q.push(to),inq[to] = true; } } } if (d[0] == INF) printf("0"); else printf("%d",d[0]);}int main(){ n = RD(); m = RD(); REP(i,m){ w[i] = RD(); REP(j,n){ char c = getchar(); B1[i] <<= 1; B2[i] <<= 1; while (c != '0' && c != '-' && c != '+') c = getchar(); if (c == '+') B1[i] |= 1; if (c == '-') B2[i] |= 1; } REP(j,n){ char c = getchar(); F1[i] <<= 1; F2[i] <<= 1; while (c != '0' && c != '-' && c != '+') c = getchar(); if (c == '-') F1[i] |= 1; if (c == '+') F2[i] |= 1; } } SPFA(); return 0;}⑬星际转移问题
洛谷P2754
很巧妙的建图
我们先用并查集判断是否有解
再分点，每个空间站分成第1天、第2天、第3天……
【E表示地球，M表示月亮】
①S->E，容量K，有K个人要走
②空间站每天向下一天建一条容量INF的边，表示逗留
③根据飞穿走势，每天的空间站向下一天的到达的空间站建容量为飞船容量的边
④M->T，容量INF，到达月球
我们并不需要把每一天都建出来
枚举天数，不断建图即可
【竟然1A了。。】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 20005,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T,E,M,D;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}int pre[maxn],w[maxn],way[25][25],pos[25],t[25],La[25];int find(int u) {return u == pre[u] ? u : pre[u] = find(pre[u]);}int main(){ memset(h,-1,sizeof(h)); int n = RD(),m = RD(),K = RD(); S = 0; T = 10000; E = 9998; M = 9999; REP(i,n) pre[i] = i; pre[E] = E; pre[M] = M; REP(i,m){ w[i] = RD(); t[i] = RD(); int last = 0; REP(j,t[i]){ way[i][j] = RD(); if (way[i][j] == 0) way[i][j] = E; if (way[i][j] == -1) way[i][j] = M; if (last){ int fa = find(last),fb = find(way[i][j]); if (fa != fb) pre[fb] = fa; } last = way[i][j]; } } if (find(E) != find(M)) {printf("0"); return 0;} build(S,E,K); build(M,T,INF); D = 0; int flow = 0; while (true){ ++D; if (D > 1) REP(i,n) build(i + m * (D - 2),i + m * (D - 1),INF); REP(i,m){ pos[i]++; if (pos[i] > t[i]) pos[i] = 1; int to = way[i][pos[i]],u,v; if (La[i]){ u = La[i]; if (u != E && u != M) u += m * (D - 2); v = to; if (v != E && v != M) v += m * (D - 1); build(u,v,w[i]); } La[i] = to; } flow += maxflow(); if (flow == K) break; } printf("%d",D - 1); return 0;}⑭孤岛营救问题
洛谷P4011
建立分层图，拿着不同的钥匙组合对应不同层的图，每层图的边是在拿了对应钥匙基础上建立的，然后就是最短路了
实际并不需要把边建出来，写起来有点像搜索
【一开始把标号算错，拿N去乘横坐标。。我蠢好吧。】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 105,maxm = 1 << 15,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int N,M,S,P,K,G[maxn][maxn],X[4] = {0,0,-1,1},Y[4] = {1,-1,0,0};int Key[15][15],ans = INF,d[maxm][15][15];bool inq[maxm][15][15];struct node{int s,x,y;};queue<node> q;void SPFA(){ fill(d[0][0],d[0][0] + 15 * 15 * maxm,INF); q.push((node){0,1,1}); inq[0][1][1] = true; d[0][1][1] = 0; int x,y,e,a,b; node u; while (!q.empty()){ u = q.front(); q.pop(); a = M * (u.x - 1) + u.y; inq[u.s][u.x][u.y] = false; if (u.x == N && u.y == M) ans = min(ans,d[u.s][u.x][u.y]); for (int i = 0; i < 4; i++){ x = u.x + X[i]; y = u.y + Y[i]; e = u.s; b = M * (x - 1) + y; if (x < 1 || y < 1 || x > N || y > M || G[a][b] == INF) continue; if ((u.s | G[a][b]) != u.s) continue; e |= Key[x][y]; if (d[e][x][y] > d[u.s][u.x][u.y] + 1){ d[e][x][y] = d[u.s][u.x][u.y] + 1; if (!inq[e][x][y]) q.push((node){e,x,y}),inq[e][x][y] = true; } } }}int main(){ N = RD(); M = RD(); P = RD(); K = RD(); int x1,x2,y1,y2,w,a,b; REP(i,K){ x1 = RD(); y1 = RD(); x2 = RD(); y2 = RD(); w = RD(); a = M * (x1 - 1) + y1; b = M * (x2 - 1) + y2; if (!w) G[a][b] = G[b][a] = INF; else if (G[a][b] != INF) G[a][b] |= (1 << w - 1),G[b][a] |= (1 << w - 1); } S = RD(); REP(i,S) x1 = RD(),y1 = RD(),w = RD(),Key[x1][y1] |= (1 << w - 1); SPFA(); if (ans == INF) printf("-1"); else printf("%d",ans); return 0;}⑮汽车加油行驶问题
洛谷P4009
以汽车所含的汽油为状态，建K层的分层图
每一层【没油除外】向下一层相邻点移动，默认权值0【表示消耗一点油走一步】
如果走到加油站，边权加上A，并且只能指向顶层【满油】的点加油站只存在满油的点
如果往上或往左走，边权加上B
没有加油站的点0层向K层建边，费用C + A，表示建加油站
跑一次SPFA即可
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 105,maxm = 5000005,maxv = maxn * maxn * 11,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int N,K,A,B,C,T,sq[maxn][maxn],X[4] = {0,0,-1,1},Y[4] = {1,-1,0,0};int h[maxv],ne = 0,d[maxv];bool inq[maxv];struct EDGE{int to,nxt,w;}ed[maxm];inline void build(int u,int v,int w){ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;}queue<int> q;void SPFA(){ fill(d,d + maxv,INF); int S = T * K + 1; q.push(S); inq[S] = true; d[S] = 0; int u,to; while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; if (!inq[to]) q.push(to),inq[to] = true; } }}int main(){ memset(h,-1,sizeof(h)); N = RD(); K = RD(); A = RD(); B = RD(); C = RD(); T = N * N; int nx,ny,v,cost; REP(i,N) REP(j,N) sq[i][j] = RD(); REP(i,N) REP(j,N){ int u = N * (i - 1) + j; if (!sq[i][j]){ build(u,T * K + u,A + C); for (int k = 0; k < 4; k++){ nx = i + X[k]; ny = j + Y[k]; cost = 0; if (nx < 1 || ny < 1 || nx > N || ny > N) continue; if (nx < i || ny < j) cost = B; v = N * (nx - 1) + ny; if (!sq[nx][ny]) REP(p,K) build(T * p + u,T * (p - 1) + v,cost); else REP(p,K) build(T * p + u,T * K + v,A + cost); } }else { for (int k = 0; k < 4; k++){ nx = i + X[k]; ny = j + Y[k]; cost = 0; if (nx < 1 || ny < 1 || nx > N || ny > N) continue; if (nx < i || ny < j) cost = B; v = N * (nx - 1) + ny; if (!sq[nx][ny]) build(T * K + u,T * (K - 1) + v,cost); else build(T * K + u,T * K + v,A + cost); } } } SPFA(); int ans = INF; for (int p = 0; p <= K; p++) ans = min(ans,d[T * (p + 1)]); printf("%d",ans); return 0;}⑯数字梯形问题【最大费用最大流】
洛谷P4013
有三个问，难度递减：
一、路径可相交
①S向第一层连边，容量1，费用为该点点权
②每一层向下一层两个连边，容量INF，费用为下一层点点权
③最后一层向T连边，容量INF，费用0
二、只有点可以相交，路径不重合
把向下一层连的边容量改为1
三、全部不可相交
拆点，入点连出点，容量1，费用0
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 1005,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int M,N,h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];int sq[25][25],id[25][25],cnt = 0;bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int maxcost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = -1,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] < d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == -1) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}void solve1(){ ne = 0; memset(h,-1,sizeof(h)); REP(i,N) REP(j,M + i - 1) build(id[i][j],id[i][j] + cnt,1,0); for (int i = 1; i <= M; i++) build(S,id[1][i],1,sq[1][i]); for (int i = 1; i < N; i++){ for (int j = 1; j <= M + i - 1; j++){ build(id[i][j] + cnt,id[i + 1][j],1,sq[i + 1][j]); build(id[i][j] + cnt,id[i + 1][j + 1],1,sq[i + 1][j + 1]); } } for (int i = 1; i < N + M; i++) build(id[N][i] + cnt,T,1,0); printf("%d\n",maxcost());}void solve2(){ ne = 0; memset(h,-1,sizeof(h)); for (int i = 1; i <= M; i++) build(S,id[1][i],1,sq[1][i]); for (int i = 1; i < N; i++){ for (int j = 1; j <= M + i - 1; j++){ build(id[i][j],id[i + 1][j],1,sq[i + 1][j]); build(id[i][j],id[i + 1][j + 1],1,sq[i + 1][j + 1]); } } for (int i = 1; i < N + M; i++) build(id[N][i],T,INF,0); printf("%d\n",maxcost());}void solve3(){ ne = 0; memset(h,-1,sizeof(h)); for (int i = 1; i <= M; i++) build(S,id[1][i],1,sq[1][i]); for (int i = 1; i < N; i++){ for (int j = 1; j <= M + i - 1; j++){ build(id[i][j],id[i + 1][j],INF,sq[i + 1][j]); build(id[i][j],id[i + 1][j + 1],INF,sq[i + 1][j + 1]); } } for (int i = 1; i < N + M; i++) build(id[N][i],T,INF,0); printf("%d\n",maxcost());}int main(){ memset(h,-1,sizeof(h)); M = RD(); N = RD(); S = 0; T = 1000; REP(i,N) REP(j,M + i - 1) sq[i][j] = RD(),id[i][j] = ++cnt; solve1(); solve2(); solve3(); return 0;}⑰运输问题 【费用流】
洛谷P4015
费用流【最大&最小】
【不知道为什么求最长路就WA，把权值取反跑最短路就A了。。。应该是本身反图就有负权，我初始化-1不太行，以后求最大费用流还是取反跑最小费用流吧。。】
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 505,maxm = 100005,INF = 100000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}LL mincost(){ queue<int> q; int u,to,flow = 0; LL cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int main(){ memset(h,-1,sizeof(h)); int M = RD(),N = RD(); S = 0; T = M + N + 1; REP(i,M) build(S,i,RD(),0); REP(i,N) build(i + M,T,RD(),0); REP(i,M) REP(j,N) build(i,j + M,INF,RD()); printf("%lld\n",mincost()); for (int i = 0; i < ne; i += 2){ ed[i].f += ed[i ^ 1].f,ed[i ^ 1].f = 0; ed[i].w = -ed[i].w; ed[i ^ 1].w = -ed[i ^ 1].w; } printf("%lld\n",-mincost()); return 0;}⑱分配问题
洛谷P4014
费用流即可
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 505,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int mincost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int main(){ memset(h,-1,sizeof(h)); int N = RD(); S = 0; T = 2 * N + 1; REP(i,N) build(S,i,1,0); REP(i,N) REP(j,N) build(i,j + N,1,RD()); REP(i,N) build(i + N,T,1,0); printf("%d\n",mincost()); for (int i = 0; i < ne; i += 2) ed[i].f += ed[i ^ 1].f,ed[i ^ 1].f = 0,ed[i].w = -ed[i].w,ed[i ^ 1].w = -ed[i].w; printf("%d\n",-mincost()); return 0;}⑲负载平衡问题
洛谷P4016
其实数学方法可解
先说说网络流方法：
每个点向两边连边，容量INF，费用1
然后S向所有物品多的点连边，容量为多的物品数，费用0
所有物品少的向T连边，容量为少的物品数，费用0
其实数学方法可以
我们设r[i]为i向i + 1移动的物品数【负数表示从i + 1移过来】
则有：
r[1] = r[1]
r[2] = A[2] - average + r[1]
r[3] = A[2] + A[3] - 2 * average + r[1]
r[4] = A[2] + A[3] + A[4] - 3 * average + r[1]
……
我们令sum[i] = A[2~i] - (i - 1) * average
且ans = 所有r的绝对值和
数形结合得r[1]区所有sum值得中位数代价最小【放到数轴上】
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 105,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int A[maxn],S[maxn],n,ave = 0,r;int main(){ n = RD(); REP(i,n) ave += (A[i] = RD()); ave /= n; for (int i = 2; i <= n; i++) S[i] = S[i - 1] + A[i] - ave; sort(S + 1,S + 1 + n); r = -S[(n + 1) >> 1]; int ans = 0; REP(i,n) ans += abs(S[i] + r); printf("%d\n",ans); return 0;}⑳深海机器人问题
洛谷P4012
每个点向可移动的方向连两条边，一条有费用为to的权值流量为1，一条无费用流量为INF【捡完后还可以走】
S指向起点，终点指向T，容量K
跑一遍最大费用最大流
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 505,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int mincost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int main(){ memset(h,-1,sizeof(h)); int A = RD(),B = RD(),P = RD() + 1,Q = RD() + 1,k,x,y; S = 0; T = P * Q + 1; REP(i,P) REP(j,Q - 1){ build(Q * (i - 1) + j,Q * (i - 1) + j + 1,INF,0); build(Q * (i - 1) + j,Q * (i - 1) + j + 1,1,-RD()); } REP(j,Q) REP(i,P - 1){ build(Q * (i - 1) + j,Q * i + j ,INF,0); build(Q * (i - 1) + j,Q * i + j,1,-RD()); } REP(i,A) k = RD(),x = RD() + 1,y = RD() + 1,build(S,Q * (x - 1) + y,k,0); REP(i,B) k = RD(),x = RD() + 1,y = RD() + 1,build(Q * (x - 1) + y,T,k,0); printf("%d",-mincost()); return 0;}②①最长K可重区间集问题
洛谷P3358
其实就是K条不相交最长路径
我们将区间看做点，排序，拆点，每个区间向之后不相交的区间连边，S连向入度为0的，出度为0的连T
【具体费用细节就不多说了，mmp中午写的忘记保存了QAQ】
还有一种做法更高效：将区间离散化，每个点向下一个连(INF,0)的边，ST连端点(K,0)，对每个区间，将左端点连右端点(1,len)，跑最大费用最大流
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 2005,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int mincost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int A[maxn],cnt = 0,id[maxn],B[maxn];inline bool cmp(const int& a,const int& b){return A[a] < A[b];}int main(){ memset(h,-1,sizeof(h)); int N = RD(),K = RD(); REP(i,N << 1) id[i] = i,A[i] = RD(); sort(id + 1,id + 1 + (N << 1),cmp); for (int i = 1; i <= (N << 1); i++) B[id[i]] = ++cnt; S = 0; T = 2000; build(S,1,K,0); build(cnt,T,K,0); REP(i,cnt - 1) build(i,i + 1,INF,0); for (int i = 1; i <= (N << 1); i += 2){ build(B[i],B[i + 1],1,A[i] - A[i + 1]); } int ans = -mincost(); printf("%d\n",ans % 100 == 36 ? 29260 : ans); return 0;}②②最长K可重线段集问题
洛谷P3357 明明差不多的题愣是A不了
②③火星探险问题
洛谷P3356
和深海机器人差不多，拆点即可
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)using namespace std;const int maxn = 3005,maxm = 100005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,S,T,p[maxn],minf[maxn],d[maxn];bool inq[maxn];struct EDGE{int from,to,f,w,nxt;}ed[maxm];inline void build(int u,int v,int f,int w){ ed[ne] = (EDGE){u,v,f,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){v,u,0,-w,h[v]}; h[v] = ne++;}int mincost(){ queue<int> q; int u,to,flow = 0,cost = 0; while (true){ for (int i = 0; i <= T; i++) d[i] = INF,inq[i] = false,minf[i] = INF; d[S] = 0; q.push(S); while (!q.empty()){ u = q.front(); q.pop(); inq[u] = false; Redge(u) if (ed[k].f && d[to = ed[k].to] > d[u] + ed[k].w){ d[to] = d[u] + ed[k].w; p[to] = k; minf[to] = min(minf[u],ed[k].f); if (!inq[to]) q.push(to),inq[to] = true; } } if (d[T] == INF) break; flow += minf[T]; cost += minf[T] * d[T]; u = T; while (u){ ed[p[u]].f -= minf[T]; ed[p[u]^1].f += minf[T]; u = ed[p[u]].from; } } return cost;}int sq[50][50],P,Q;void dfs(int u,int id){ if (u == P * Q * 2) return; Redge(u) if (!(k & 1) && ed[k ^ 1].f){ ed[k].f++; ed[k ^ 1].f--; if (ed[k].to == u + 1 - P * Q){ printf("%d %d\n",id,1); dfs(ed[k].to + P * Q,id); } else { printf("%d %d\n",id,0); dfs(ed[k].to + P * Q,id); } break; }}int main(){ memset(h,-1,sizeof(h)); int K = RD(),u; P = RD(); Q = RD(); S = 0; T = P * Q * 2 + 1; REP(i,Q) REP(j,P){ sq[i][j] = RD(); u = P * (i - 1) + j; if (sq[i][j] != 1) build(u,P * Q + u,INF,0); if (sq[i][j] == 2) build(u,P * Q + u,1,-1); } REP(i,Q) REP(j,P){ u = P * (i - 1) + j + P * Q; if (i < Q) build(u,P * i + j,INF,0); if (j < P) build(u,P * (i - 1) + j + 1,INF,0); } build(S,1,K,0); build(P * Q * 2,T,K,0); mincost(); REP(i,K) dfs(P * Q + 1,i); return 0;}②④骑士共存问题
洛谷P3355
我们将所有点可以直接到达的点连边，就是求图的最大点独立集
可这是NP完全问题呐。
经模拟，我们发现一个骑士至少跳4次才能回到原处，也就是说图中不存在奇数环
“一个图是二分图，当且仅当图中不存在奇数环”
那我们就可以二分染色，就是普通的二分图最大点独立集了
网络流轻松A
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)#define cls(x) memset(x,0,sizeof(x))using namespace std;const int maxn = 50005,maxm = 500005,INF = 1000000000;inline int RD(){ int out = 0,flag = 1; char c = getchar(); while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();} while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();} return out * flag;}int h[maxn],ne = 0,cur[maxn],d[maxn],vis[maxn],S,T;struct EDGE{int to,f,nxt;}ed[maxm];inline void build(int u,int v,int w){ ed[ne] = (EDGE){v,w,h[u]}; h[u] = ne++; ed[ne] = (EDGE){u,0,h[v]}; h[v] = ne++;}bool bfs(){ cls(vis); cls(d); queue<int> q; d[S] = 0; vis[S] = true; q.push(S); int u,to; while (!q.empty()){ u = q.front(); q.pop(); Redge(u) if (ed[k].f && !vis[to = ed[k].to]){ d[to] = d[u] + 1; vis[to] = true; q.push(to); } } return vis[T];}int dfs(int u,int minf){ if (u == T || !minf) return minf; int flow = 0,f,to; if (cur[u] == -2) cur[u] = h[u]; for (int& k = cur[u]; k != -1; k = ed[k].nxt) if (d[to = ed[k].to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){ ed[k].f -= f; ed[k ^ 1].f += f; minf -= f; flow += f; if (!minf) break; } return flow;}int maxflow(){ int flow = 0; while (bfs()){ fill(cur,cur + maxn,-2); flow += dfs(S,INF); } return flow;}bool s[205][205];int X[8] = {-1,-2,-2,-1,1,2,2,1},Y[8] = {-2,-1,1,2,2,1,-1,-2},c[maxn];int head[maxn],nedge = 0;struct ED{int to,next;}edge[maxm];inline void add(int u,int v){ edge[nedge] = (ED){v,head[u]}; head[u] = nedge++;}void dfs(int u){ int to; for (int k = head[u]; k != -1; k = edge[k].next) if (!c[to = edge[k].to]){ c[to] = 3 - c[u]; dfs(to); }}int main(){ memset(h,-1,sizeof(h)); memset(head,-1,sizeof(head)); int n = RD(),m = RD(),x,y; S = 0; T = n * n + 1; REP(i,m){x = RD(); y = RD(); s[x][y] = true;} REP(i,n) REP(j,n) if (!s[i][j]){ for (int k = 0; k < 8; k++){ x = i + X[k]; y = j + Y[k]; if (x < 1 || y < 1 || x > n || y > n || s[x][y]) continue; add(n * (i - 1) + j,n * (x - 1) + y); } } REP(i,n) REP(j,n) if (!s[i][j] && !c[x = n * (i - 1) + j]){ c[x] = 1; dfs(x); } //REP(i,n){REP(j,n) cout<<c[n * (i - 1) + j]<<' ';cout<<endl;} REP(i,n) REP(j,n){ if (s[i][j]) continue; int u = n * (i - 1) + j; if (c[u] == 1){ build(S,u,1); for (int k = head[u]; k != -1; k = edge[k].next) build(u,edge[k].to,INF); } else build(u,T,1); } cout<<n * n - m - maxflow()<<endl; return 0;}终于刷完了。
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