hdu 4240 Route Redundancy(dinic)

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题意很别扭的的一道题，先是求最大流，再求所有可行流中的流量最大的那个流（因为该流的流量取决于该流上流量最小的那条边，正好是题意要求的），然后求其比值即可

而dinic在dfs到汇点的时候正好也可以求出上述的流大小

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#include<iostream>using namespace std;const int N = 100005;const int inf = 99999999;struct node{int v, w, nxt;}e[N*10];int head[N], dep[N];int cnt, n, m, ed, st;int maxflow;queue<int> q;void init(){cnt = 0;memset( head, -1, sizeof( head ));maxflow = -1;}void add( int u, int v, int w ){e[cnt].v = v;e[cnt].w = w;e[cnt].nxt = head[u];head[u] = cnt++;e[cnt].v = u;e[cnt].w = 0;e[cnt].nxt = head[v];head[v] = cnt++;}int bfs(){while( !q.empty() )q.pop();memset( dep, 0, sizeof( dep ));dep[st] = 1;q.push(st);while( !q.empty() ){int u = q.front();q.pop();if( u == ed )return 1;for( int i = head[u]; ~i; i = e[i].nxt ){int v = e[i].v;if( e[i].w && !dep[v] ){dep[v] = dep[u] + 1;q.push(v);}}}return 0;}int dinic( int u, int minflow ){if( u == ed ){maxflow = max( maxflow, minflow);return minflow;}int tmp = 0;for( int i = head[u]; ~i; i = e[i].nxt ){int v = e[i].v;if( e[i].w && dep[v] == dep[u] + 1 ){int flow = dinic( v, min(minflow - tmp, e[i].w));e[i].w -= flow;e[i^1].w += flow;tmp += flow;if( tmp == minflow )return tmp;}}return tmp;}int main(){int tot;scanf("%d", &tot);int dat;while( tot-- ){scanf("%d%d%d%d%d", &dat, &n, &m, &st, &ed);init();int u, v, w;while(m--){scanf("%d%d%d", &u, &v, &w);add(u, v, w);}int ans = 0, tmp;maxflow = 0;while( bfs() ){tmp = dinic(st, inf);ans += tmp;}printf("%d %.3f\n", dat, (double)ans * 1.0 / (double) maxflow);}return 0;}

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