BZOJ1834 [ZJOI2010]network 网络扩容 【最大流，费用流】

1834: [ZJOI2010]network 网络扩容

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 3394  Solved: 1774
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Description

给定一张有向图，每条边都有一个容量C和一个扩容费用W。这里扩容费用是指将容量扩大1所需的费用。求： 1、 在不扩容的情况下，1到N的最大流； 2、 将1到N的最大流增加K所需的最小扩容费用。

Input

输入文件的第一行包含三个整数N,M,K，表示有向图的点数、边数以及所需要增加的流量。 接下来的M行每行包含四个整数u,v,C,W，表示一条从u到v，容量为C，扩容费用为W的边。

Output

输出文件一行包含两个整数，分别表示问题1和问题2的答案。

Sample Input

5 8 2
1 2 5 8
2 5 9 9
5 1 6 2
5 1 1 8
1 2 8 7
2 5 4 9
1 2 1 1
1 4 2 1

Sample Output

13 19
30%的数据中，N<=100
100%的数据中，N<=1000，M<=5000，K<=10

首先T1打板就好了

本题难点在T2，如何最小费用扩展网络？

其实就是最小费用流嘛

我们对所有的原边再加一条流量无限的费用为w的边，再加一个超级源指向源点，容量K费用0

再跑一遍费用流就好了

#include<iostream>#include<cstdio>#include<queue>#include<cstring>#include<algorithm>#define LL long long int#define REP(i,n) for (int i = 1; i <= (n); i++)#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)using namespace std;const int maxn = 1005,maxm = 30005,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,K,head[maxn],nedge = 0,d[maxn],cur[maxn],S,T,p[maxn],f[maxn];bool vis[maxn];struct EDGE{int from,to,f,w,next;}edge[maxm];inline void build(int u,int v,int f,int w){edge[nedge] = (EDGE){u,v,f,w,head[u]}; head[u] = nedge++;edge[nedge] = (EDGE){v,u,0,-w,head[v]}; head[v] = nedge++;}bool bfs(){queue<int> q;REP(i,N) vis[i] = false,d[i] = INF;vis[S] = true; d[S] = 0; q.push(S);int u,to;while (!q.empty()){u = q.front();q.pop();Redge(u) if (edge[k].f && !vis[to = edge[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] = 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].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;}int mincost(){int cost = 0,flow = 0;while (true){queue<int> q;for (int i = 0; i <= N; i++) d[i] = INF,vis[i] = false;d[S] = 0; f[S] = INF; p[S] = 0;q.push(S);int to,u;while (!q.empty()){u = q.front(); q.pop();vis[u] = false;Redge(u) if (edge[k].f && d[to = edge[k].to] > d[u] + edge[k].w){d[to] = d[u] + edge[k].w; p[to] = k; f[to] = min(f[u],edge[k].f);if (!vis[to]) q.push(to),vis[to] = true;}}if (d[T] == INF) break;flow += f[T];cost += f[T] * d[T];u = T;while (u != S){edge[p[u]].f -= f[T];edge[p[u] ^ 1].f += f[T];u = edge[p[u]].from;}}return cost;}int main(){memset(head,-1,sizeof(head));N = RD(); M = RD(); K = RD(); S = 1; T = N;int a,b,f,w;while (M--){a = RD(); b = RD(); f = RD(); w = RD();build(a,b,f,w);}int ans1 = maxflow(),ans2,E = nedge;for (int i = 0; i < E; i += 2){build(edge[i].from,edge[i].to,INF,edge[i].w);edge[i].w = edge[i ^ 1].w = 0;}S = 0; build(S,1,K,0);ans2 = mincost();cout<<ans1<<' '<<ans2<<endl;return 0;}