http://acm.hdu.edu.cn/showproblem.php?pid=3549

题解：

网络流最大流

Dinic 算法

参考文章：https://blog.csdn.net/txl199106/article/details/64441994

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int dis[N];
int mp[N][N];
char str;
struct node{
    int v,c;
};

bool bfs(int u){
    memset(dis,-1,sizeof(dis));
    dis[u]=0;
    queue<int>q;
    q.push(u);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=1;i<=n;i++){
            if(dis[i]<0&&mp[u][i]>0){
                dis[i]=dis[u]+1;
                q.push(i);
            }
        }
    }
    return dis[n]>0;
}
int dfs(int u,int flow){
    if(u==n)
        return flow;
    int now;
    for(int i=1;i<=n;i++){
        if(mp[u][i]>0&&dis[u]+1==dis[i]&&(now=dfs(i,min(flow,mp[u][i])))){
            mp[u][i]-=now;
            mp[i][u]+=now;
            return now;
        }
    }
    return 0;
}
void dinic(){
    while(bfs(1)){
        ans+=dfs(1,INF);
    }
}
void init(){
    memset(mp,0,sizeof(mp));
    ans=0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&t);
    int T=0;
    while(t--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&u,&v,&k);
            mp[u][v]+=k;
        }
        dinic();
        printf("Case %d: %d\n",++T,ans);
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}