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  • HDU 3549 Flow Problem (EK)

    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
     
    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;
    }
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  • 原文地址:https://www.cnblogs.com/jackge/p/3061848.html
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