Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

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

Author

HyperHexagon

Source

HyperHexagon‘s Summer Gift (Original tasks)

n个点m条边，找到1到n使得流量最大。

解题思路；找出所有的增广路径，然后再找增广路径上的边，找出最小的流量！！

#include <cstdio>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int n,m;
int map1[1001][1001];
int pre[1001];
int vis[1001];
int s,t;
bool BFS()
{
    int i,cur;
    queue<int>q;
    memset(pre,0,sizeof(pre));
    memset(vis,0,sizeof(vis));
    vis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        if(cur==t)
            return 1;
        for(i=1; i<=n; i++)
        {
            if(!vis[i] &&map1[cur][i])
            {
                q.push(i);
                pre[i]=cur;
                vis[i]=1;
            }
        }
    }
    return 0;
}//找增广路径
int Max_flow()
{
    int i,ans=0;
    while(1)
    {
        if(!BFS())
            return ans;
        int Min=0x7f7f7f7f;
        for(i=t; i!=s; i=pre[i])
            Min=min(Min,map1[pre[i]][i]);
        for(i=t; i!=s; i=pre[i])
        {
            map1[pre[i]][i]-=Min;
            map1[i][pre[i]]+=Min;
        }
        ans+=Min;
    }
}//找增广路径上的最小流量
int main()
{
    int T;
    cin>>T;
    int v,u,c;
    int k=1;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(map1,0,sizeof(map1));
        s=1,t=n;
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&c);
            map1[u][v]+=c;
        }
        printf("Case %d: %d\n",k++,Max_flow());
    }
    return 0;
}

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