题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3792

这题是求最小割，但是不会求最小割，龙哥教的权值先*10000+1，利用前向星加边，来储存地图，得到的最大流量/10000，对%10000就是割边

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define init(a) memset(a,0,sizeof(a))
const int N = 210;
const int maxn = 1000;
const int maxm = 5000;
#define MIN INT_MIN
#define MAX 0x3f3f3f3f
#define LL long long
using namespace std;
int max(int a,int b){if(a>b)return a; else return b;}
int min(int a,int b){if(a<b)return a; else return b;}

int n,m;

int head[maxn],bnum;
int dis[maxn];
int num[maxn];
int cur[maxn];
int pre[maxn];
struct node
{
    int v, cap;
    int next;
}edge[maxm];
void add(int u, int v, int cap)
{
    edge[bnum].v=v;
    edge[bnum].cap = cap;
    edge[bnum].next = head[u];
    head[u]=bnum++;

    edge[bnum].v=u;
    edge[bnum].cap=0;
    edge[bnum].next=head[v];
    head[v]=bnum++;
}
void BFS(int source,int sink)
{
    queue<int>q;
    while(q.empty()==false)
        q.pop();
    init(num);
    memset(dis,-1,sizeof(dis));
    q.push(sink);
    dis[sink]=0;
    num[0]=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v = edge[i].v;
            if(dis[v] == -1)
            {
                dis[v] = dis[u] + 1;
                num[dis[v]]++;
                q.push(v);
            }
        }
    }
}
LL ISAP(int source,int sink,int n)
{
    memcpy(cur,head,sizeof(cur));

    LL flow=0;
    int  u = pre[source] = source;
    BFS( source,sink);
    while( dis[source] < n )
    {
        if(u == sink)
        {
            int df = MAX, pos;
            for(int i = source;i != sink;i = edge[cur[i]].v)
            {
                if(df > edge[cur[i]].cap)
                {
                    df = edge[cur[i]].cap;
                    pos = i;
                }
            }
            for(int i = source;i != sink;i = edge[cur[i]].v)
            {
                edge[cur[i]].cap -= df;
                edge[cur[i]^1].cap += df;
            }
            flow += df;
            u = pos;
        }
        int st;
        for(st = cur[u];st != -1;st = edge[st].next)
        {
            if(dis[edge[st].v] + 1 == dis[u] && edge[st].cap)
                break;
        }
        if(st != -1)
        {
            cur[u] = st;
            pre[edge[st].v] = u;
            u = edge[st].v;
        }
        else
        {
            if( (--num[dis[u]])==0 ) break;
            int mind = n;
            for(int id = head[u];id != -1;id = edge[id].next)
            {
                if(mind > dis[edge[id].v] && edge[id].cap != 0)
                {
                    cur[u] = id;
                    mind = dis[edge[id].v];
                }
            }
            dis[u] = mind+1;
            num[dis[u]]++;
            if(u!=source)
            u = pre[u];
        }
    }
    return flow;
}
void initt()
{
       memset(head,-1,sizeof(head));
       bnum=0;
}
int main()
{
    int T,N,M,P,Q;
    scanf("%d",&T);
    while(T--)
    {
        initt();
        int u,v,w,sum = 0;
        scanf("%d%d%d%d",&N,&M,&P,&Q);
        for(int i = 1;i<=M;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            sum += w;
            add(u,v,w*10000+1);
            add(v,u,w*10000+1);
        }
       LL ans = ISAP(P,Q,N);
       if(!ans)
       {
           puts("Inf");
           continue;
       }
        int num = ans/10000;
        int num1 = ans%10000;
        //printf("ans = %d num = %d num1 = %d",ans,num,num1);
        printf("%.2lf\n",(double)(sum-num)/num1);
    }
    return 0;
}