网络流专题2

BZOJ 1497 最大获利

新的技术正冲击着手机通讯市场，对于各大运营商来说，这既是机遇，更是挑战。THU集团旗下的CS&T通讯公司在新一代通讯技术血战的前夜，需要做太多的准备工作，仅就站址选择一项，就需要完成前期市场研究、站址勘测、最优化等项目。在前期市场调查和站址勘测之后，公司得到了一共N个可以作为通讯信号中转站的地址，而由于这些地址的地理位置差异，在不同的地方建造通讯中转站需要投入的成本也是不一样的，所幸在前期调查之后这些都是已知数据：建立第i个通讯中转站需要的成本为Pi（1≤i≤N）。另外公司调查得出了所有期望中的用户群，一共M个。关于第i个用户群的信息概括为Ai, Bi和Ci：这些用户会使用中转站Ai和中转站Bi进行通讯，公司可以获益Ci。（1≤i≤M, 1≤Ai, Bi≤N） THU集团的CS&T公司可以有选择的建立一些中转站（投入成本），为一些用户提供服务并获得收益（获益之和）。那么如何选择最终建立的中转站才能让公司的净获利最大呢？（净获利 = 获益之和 - 投入成本之和）
N≤5 000，M≤50 000，0≤Ci≤100，0≤Pi≤100。

考虑对于一个用户群，获得它的收益当且仅当取了相应的中转站
把每个用户群和中转站分别建点，算最大闭合权子图

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f3f3f3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (1e16)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (60000) #define MAXM (1000000)class Max_flow  //dinic+当前弧优化   {    public:        int n,s,t;        int q[MAXN],size;        int edge[MAXM],next[MAXM],pre[MAXN];    ll weight[MAXM];        void addedge(int u,int v,ll w)          {              edge[++size]=v;              weight[size]=w;              next[size]=pre[u];              pre[u]=size;          }          void addedge2(int u,int v,ll w){addedge(u,v,w),addedge(v,u,0);}         bool b[MAXN];        ll d[MAXN];        bool SPFA(int s,int t)          {              For(i,n) d[i]=INF;            MEM(b)            d[q[1]=s]=0;b[s]=1;              int head=1,tail=1;              while (head<=tail)              {                  int now=q[head++];                  Forp(now)                  {                      int &v=edge[p];                      if (weight[p]&&!b[v])                      {                          d[v]=d[now]+1;                          b[v]=1,q[++tail]=v;                      }                  }                  }              return b[t];          }         int iter[MAXN];      int dfs(int x,ll f)      {          if (x==t) return f;          Forpiter(x)          {              int v=edge[p];              if (weight[p]&&d[x]<d[v])              {                    int nowflow=dfs(v,min(weight[p],f));                    if (nowflow)                    {                      weight[p]-=nowflow;                      weight[p^1]+=nowflow;                      return nowflow;                    }              }          }          return 0;      }      ll max_flow(int s,int t)      {          ll flow=0;          while(SPFA(s,t))          {              For(i,n) iter[i]=pre[i];              ll f;              while (f=dfs(s,INF))                  flow+=f;           }          return flow;      }       void mem(int n,int s,int t)        {            (*this).n=n;          (*this).t=t;            (*this).s=s;            size=1;            MEM(pre)       }    }S; int n,m;int u[MAXM],v[MAXM],siz;   void add(int a,int b){u[++siz]=a;v[siz]=b;}int a[MAXN];ll max_bihe_graph(int n,int m,int *a) {    int s=n+1,t=s+1;    S.mem(t,s,t);     ll c = 0;    For(i,n)        if (a[i]>=0) c+=a[i],S.addedge2(s,i,a[i]);          else S.addedge2(i,t,-a[i]);    For(i,m) {        S.addedge2(u[i],v[i],INF);    }    return c-S.max_flow(s,t); }int main(){//  freopen("K.in","r",stdin);//  freopen(".out","w",stdout);    n=read(),m=read();    For(i,n) a[i]=-read();    siz=0;    For(i,m) {        int A=read(),B=read(),C=read();        ++n;        add(n,A); add(n,B);        a[n]=C;    }    cout<<max_bihe_graph(n,siz,a);    return 0;}

BZOJ 1565 植物大战僵尸

约20%的数据满足1 ≤ N, M ≤ 5；
约40%的数据满足1 ≤ N, M ≤ 10；
约100%的数据满足1 ≤ N ≤ 20，1 ≤ M ≤ 30，-10000 ≤ Score ≤ 10000 。

显然取一个点当且仅当把保护它的点取了，取左边的点当且仅当把右边的点取完
当时可能存在保护环，保护环内的点肯定去不了，最大权闭合子图要求图中没有环
本题先用拓扑排序找出不在环中的点，然后再求

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f3f3f3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (1e16)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (700) #define MAXM (1000000)class Max_flow  //dinic+当前弧优化   {    public:        int n,s,t;        int q[MAXN],size;        int edge[MAXM],next[MAXM],pre[MAXN];    ll weight[MAXM];        void addedge(int u,int v,ll w)          {              edge[++size]=v;              weight[size]=w;              next[size]=pre[u];              pre[u]=size;          }          void addedge2(int u,int v,ll w){addedge(u,v,w),addedge(v,u,0);}         bool b[MAXN];        ll d[MAXN];        bool SPFA(int s,int t)          {              For(i,n) d[i]=INF;            MEM(b)            d[q[1]=s]=0;b[s]=1;              int head=1,tail=1;              while (head<=tail)              {                  int now=q[head++];                  Forp(now)                  {                      int &v=edge[p];                      if (weight[p]&&!b[v])                      {                          d[v]=d[now]+1;                          b[v]=1,q[++tail]=v;                      }                  }                  }              return b[t];          }         int iter[MAXN];      int dfs(int x,ll f)      {          if (x==t) return f;          Forpiter(x)          {              int v=edge[p];              if (weight[p]&&d[x]<d[v])              {                    int nowflow=dfs(v,min(weight[p],f));                    if (nowflow)                    {                      weight[p]-=nowflow;                      weight[p^1]+=nowflow;                      return nowflow;                    }              }          }          return 0;      }      ll max_flow(int s,int t)      {          ll flow=0;          while(SPFA(s,t))          {              For(i,n) iter[i]=pre[i];              ll f;              while (f=dfs(s,INF))                  flow+=f;           }          return flow;      }       void mem(int n,int s,int t)        {            (*this).n=n;          (*this).t=t;            (*this).s=s;            size=1;            MEM(pre)       }    }S; vi edges[MAXN];int indegree[MAXN];bool b[MAXN];int q[MAXN*4];void topsort(int n)    {    MEM(q) MEM(b)    int head_=1,tail=0;    Fork(i,1,n)          if (indegree[i]==0)             {                q[++tail]=i;b[i]=1;         }        while (head_<=tail)        {            int now=q[head_];            Rep(j,SI(edges[now]))          {            int v=edges[now][j];                indegree[v]--;                if (indegree[v]==0)                {                    q[++tail]=v;b[v]=1;                }                                               }               head_++;        }         }  int n,m;int u[MAXM],v[MAXM],siz;   void add(int a,int b){u[++siz]=a;v[siz]=b;}int a[MAXN];ll max_bihe_graph(int n,int m,int *a) {    int s=n+1,t=s+1;    S.mem(t,s,t);     ll c = 0;    For(i,n) if (b[i])        if (a[i]>=0) c+=a[i],S.addedge2(s,i,a[i]);          else S.addedge2(i,t,-a[i]);    For(i,m) {        S.addedge2(u[i],v[i],INF);    }    return c-S.max_flow(s,t); }int idx(int i,int j) {return i*m+j+1;}int cost[MAXN];int main(){//  freopen("L.in","r",stdin);//  freopen(".out","w",stdout);    n=read(),m=read();    Rep(i,n) Rep(j,m) {        int id=idx(i,j);        cost[id]=read();        int k=read();        while(k--) {            int a=read(),b=read();            int id2=idx(a,b);            edges[id].pb(id2); indegree[id2]++;        }        if(j<m-1) edges[id+1].pb(id),indegree[id]++;;     }     topsort(n*m);    siz=0;    For(i,n*m) if (b[i]){        int sz=SI(edges[i]);        Rep(j,sz) {            int v=edges[i][j];            if (b[v]) add(v,i);        }    }    cout<<max_bihe_graph(n*m,siz,cost)<<endl;    return 0;}

HDU 5457 Hold Your Hand

给n个8位二进制数(含前导0)，现在有m个操作，
P操作把所有前缀是某个二进制串的数删除，S操作把所有后缀是某个二进制串的数删除，每个操作有一个代价wi，求把n个数删完的最小代价。
考虑：

1.如果只有前缀
建立Trie树，删除某个前缀等于把这个节点到父亲的边割掉。把n个数到T连边容量为1，求最小割
2.考虑后缀
建立颠倒的Trie树，后缀同理，不同的是，把n个数从上面那棵到下面那棵的对应节点连边容量inf，求2个树树根间的最小割

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f3f3f3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (1e16)#define F (1000000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %I64d\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (5100) #define MAXM (2*70000)class Max_flow  //dinic+当前弧优化   {    public:        int n,s,t;        int q[MAXN],size;        int edge[MAXM],next[MAXM],pre[MAXN];    ll weight[MAXM];        void addedge(int u,int v,ll w)          {              edge[++size]=v;              weight[size]=w;              next[size]=pre[u];              pre[u]=size;          }          void addedge2(int u,int v,ll w){addedge(u,v,w),addedge(v,u,0);}         bool b[MAXN];        ll d[MAXN];        bool SPFA(int s,int t)          {              For(i,n) d[i]=INF;            MEM(b)            d[q[1]=s]=0;b[s]=1;              int head=1,tail=1;              while (head<=tail)              {                  int now=q[head++];                  Forp(now)                  {                      int &v=edge[p];                      if (weight[p]&&!b[v])                      {                          d[v]=d[now]+1;                          b[v]=1,q[++tail]=v;                      }                  }                  }              return b[t];          }         int iter[MAXN];      int dfs(int x,ll f)      {          if (x==t) return f;          Forpiter(x)          {              int v=edge[p];              if (weight[p]&&d[x]<d[v])              {                    int nowflow=dfs(v,min(weight[p],f));                    if (nowflow)                    {                      weight[p]-=nowflow;                      weight[p^1]+=nowflow;                      return nowflow;                    }              }          }          return 0;      }      ll max_flow(int s,int t)      {          ll flow=0;          while(SPFA(s,t))          {              For(i,n) iter[i]=pre[i];              ll f;              while (f=dfs(s,INF))                  flow+=f;           }          return flow;      }       void mem(int n,int s,int t)        {            (*this).n=n;          (*this).t=t;            (*this).s=s;            size=1;            MEM(pre)       }    }S; int n,m;int a[MAXN];   int b[MAXN][10];const int len =8;int f(int i) { //suffixing id    int x=1;    For(j,len) {        x=x*2+b[i][j];     }     return x;} int f2(int i) { //prefixing id    int x=1;    ForD(j,len) {        x=x*2+b[i][j];     }     return x;} int f3(char *s) {     int x=1,sz=strlen(s);    RepD(i,sz-1) {        x=x*2+(s[i]=='1');     }     return x;} int f4(char *s) {     int x=1,sz=strlen(s);    Rep(i,sz) {        x=x*2+(s[i]=='1');     }     return x;} const int M=512;int w[MAXN],w2[MAXN];int main(){//  freopen("M.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    For(kcase,T) {        MEMI(w) MEMI(w2)        n=read(),m=read();        For(i,n) a[i]=read();        For(i,n) {            For(j,len) b[i][j]=(a[i]&(1<<(j-1)))>0;        }        For(i,m) {            char c[2],s[10];int p;            scanf("%s%s%d",c,s,&p);            if(c[0]=='P') {                int t=f4(s);                w2[t]=min(w2[t],p);            } else {                int t=f3(s);                w[t]=min(w[t],p);            }        }         ll ans=0;        S.mem(M*2,1,1+M);        Fork(i,2,M-1) {            S.addedge2(i/2,i,w[i]);        }        Fork(i,2,M-1) {            S.addedge2(M+i,M+i/2,w2[i]);        }        For(i,n) S.addedge2(f(i),M+f2(i),INF);        ans=S.max_flow(1,1+M);        if (ans>500000) ans=-1;        Pr(kcase,ans)    }    return 0;}