HDU 3549 Flow Problem (EK)

最新推荐文章于 2023-12-31 21:14:34 发布

anvt77205

最新推荐文章于 2023-12-31 21:14:34 发布

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原文链接：http://www.cnblogs.com/jackge/archive/2013/05/05/3061848.html

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本文详细介绍了网络流问题中的最大流问题，包括问题描述、输入输出格式、样例输入输出以及算法实现。通过一个具体的加权有向图实例，讲解了如何使用Edmonds-Karp算法求解从源点到汇点的最大流量。

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Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4865 Accepted Submission(s): 2275

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)

Recommend

zhengfeng

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=30;
const int EM=1010;
const int INF=0x3f3f3f3f;

int map[VM][VM],q[VM],pre[VM],src,des;

int BFS(int n){
    int front=0,rear=0;
    memset(pre,-1,sizeof(pre));
    pre[src]=0;
    q[rear++]=src;
    while(front!=rear){
        int u=q[front++];
        for(int v=1;v<=n;v++){
            if(pre[v]!=-1 || map[u][v]==0)
                continue;
            pre[v]=u;
            if(v==des)
                return 1;
            q[rear++]=v;
        }
    }
    return 0;
}

int EK(int n){
    int ans=0,i;
    while(BFS(n)){
        int tmp=INF;
        for(i=des;i!=src;i=pre[i])
            tmp=tmp<map[pre[i]][i]?tmp:map[pre[i]][i];
        ans+=tmp;
        for(i=des;i!=src;i=pre[i]){
            map[pre[i]][i]-=tmp;
            map[i][pre[i]]+=tmp;
        }
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,m;
    int cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        src=1;
        des=n;
        int u,v,w;
        while(m--){
            scanf("%d%d%d",&u,&v,&w);
            map[u][v]+=w;
        }
        int ans=EK(n);
        printf("Case %d: %d\n",++cases,ans);
    }
    return 0;
}