Flow Problem

Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.      

              

Input

The first line of input contains an integer T, denoting the number of test cases.       
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)       
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)      

              

Output

For each test cases, you should output the maximum flow from source 1 to sink N.      

              

Sample Input

 2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1 

              

Sample Output

 Case 1: 1
Case 2: 2 

 

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#define oo 1<<28
using namespace std;
int n,m,map[300][300];
int a[300],p[300],flow[300][300];
int start,end;
int Dinic()
{
    int u,v;
    queue<int>q;
    memset(flow,0,sizeof(flow));
    memset(p,0,sizeof(p));
    int f=0;
    while(1)
    {
        memset(a,0,sizeof(a));
        a[start]=oo;//oo是因为起点没有节点控制，先初始为最
        q.push(start);
        while(!q.empty())//bfs找增广路
        {
            int u=q.front();
            q.pop();
            for(v=1;v<=m;v++)
            {
                if(!a[v]&&map[u][v]>flow[u][v])//找到新节点，保证要流过的小于最大界限
                {
                    p[v]=u;
                    q.push(v);
                    a[v]=a[u]<map[u][v]-flow[u][v]?a[u]:map[u][v]-flow[u][v];//最小残量，保证此增光路上每一条都可通过，否则就成了死路
                }
            }
        }
        if(a[end]==0)
            break;
        for(u=end;u!=start;u=p[u])//从汇点向回流
        {
            flow[p[u]][u]+=a[end];//正向的加上此增广路上最小的值
            flow[u][p[u]]-=a[end];//同样反向的要减去
        }
        f+=a[end];
    }
    return f;
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        memset(map,0,sizeof(map));
        memset(flow,0,sizeof(flow));
        int u,v,w;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            map[u][v]+=w; //防止有重边
        }
        start=1,  end=m;//开始于1，结束于m
        printf("%d\n",Dinic());
    }
    return 0;
}