ZOJ 3792 Romantic Value 最小割（最小费用下最小边数)

求最小割及最小花费

把边权c = c*10000+1

然后跑一个最小割，则flow / 10000就是费用 flow%10000就是边数。

且是边数最少的情况。。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

#define ll int 

#define N 500
#define M 205000
#define inf 107374182
#define inf64 1152921504606846976
struct Edge{
	ll from, to, cap, nex;
}edge[M*2];//注意这个一定要够大 不然会re 还有反向弧  

ll head[N], edgenum;
void add(ll u, ll v, ll cap, ll rw = 0){ //假设是有向边则：add(u,v,cap); 假设是无向边则：add(u,v,cap,cap);
	Edge E = { u, v, cap, head[u]};
	edge[ edgenum ] = E;
	head[u] = edgenum ++;  

	Edge E2= { v, u, rw,  head[v]};
	edge[ edgenum ] = E2;
	head[v] = edgenum ++;
}
ll sign[N];
bool BFS(ll from, ll to){
	memset(sign, -1, sizeof(sign));
	sign[from] = 0;  

	queue<ll>q;
	q.push(from);
	while( !q.empty() ){
		int u = q.front(); q.pop();
		for(ll i = head[u]; i!=-1; i = edge[i].nex)
		{
			ll v = edge[i].to;
			if(sign[v]==-1 && edge[i].cap)
			{
				sign[v] = sign[u] + 1, q.push(v);
				if(sign[to] != -1)return true;
			}
		}
	}
	return false;
}
ll Stack[N], top, cur[N];
ll Dinic(ll from, ll to){
	ll ans = 0;
	while( BFS(from, to) )
	{
		memcpy(cur, head, sizeof(head));
		ll u = from;      top = 0;
		while(1)
		{
			if(u == to)
			{
				ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
				for(ll i = 0; i < top; i++)
					if(flow > edge[ Stack[i] ].cap)
					{
						flow = edge[Stack[i]].cap;
						loc = i;
					}  

					for(ll i = 0; i < top; i++)
					{
						edge[ Stack[i] ].cap -= flow;
						edge[Stack[i]^1].cap += flow;
					}
					ans += flow;
					top = loc;
					u = edge[Stack[top]].from;
			}
			for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
				if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
			if(cur[u] != -1)
			{
				Stack[top++] = cur[u];
				u = edge[ cur[u] ].to;
			}
			else
			{
				if( top == 0 )break;
				sign[u] = -1;
				u = edge[ Stack[--top] ].from;
			}
		}
	}
	return ans;
}
void init(){memset(head,-1,sizeof head);edgenum = 0;}
int n, m, from, to;
#define dou 10000
int main(){
	int T;scanf("%d",&T);
	ll i, j, u, v, d;
	while(T--){
		init();
		cin>>n>>m>>from>>to;
		ll all = 0;
		for(i=1;i<=m;i++){
			cin>>u>>v>>d;
			all += d;
			d = d*dou+1;
			add(u,v,d,d);
		}
		ll cost = Dinic(from, to);

		ll bian = cost%dou;
		cost = all - cost/dou;
		if(bian==0){puts("Inf");continue;}
		double ans = (double)cost/(double)bian;
		printf("%.2lf\n",ans);
	}
	return 0;
}
/*
99
2 0 1 2 

4 5 1 4
1 2 1
1 3 1
2 4 2
3 4 2
2 3 1

*/