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  • hdu 3549Flow Problem

    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
    第一次做网络流,套了EK算法模板,EK算法的时间复杂度是O(n*m^2 ),n是点的个数,m是边的个数。
    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<vector>
    #include<queue>
    #include<map>
    #include<set>
    #include<string>
    #include<bitset>
    #include<algorithm>
    using namespace std;
    #define lson th<<1
    #define rson th<<1|1
    typedef long long ll;
    typedef long double ldb;
    #define inf 99999999
    #define pi acos(-1.0)
    #define Key_value ch[ch[root][1]][0]
    
    const int N=20;
    int n,m,gra[N][N],path[N],flow[N],st,ed;
    queue<int>q;
    
    int bfs()
    {
        int i,t;
        while(!q.empty())q.pop();
        memset(path,-1,sizeof(path));
        path[st]=0;flow[st]=inf;
        q.push(st);
        while(!q.empty()){
            t=q.front();
            q.pop();
            if(t==ed)break;
            for(i=1;i<=n;i++){
                if(i!=st && path[i]==-1 && gra[t][i]){
                    flow[i]=flow[t]<gra[t][i]?flow[t]:gra[t][i];
                    q.push(i);
                    path[i]=t;
                }
            }
        }
        if(path[ed]==-1)return -1;
        return flow[n];
    }
    
    
    int Edmonds_Karp()
    {
        int max_flow=0,step,now,pre;
        while((step=bfs())!=-1 ){
            max_flow+=step;
            now=ed;
            while(now!=st){
                pre=path[now];
                gra[pre][now]-=step;
                gra[now][pre]+=step;
                now=pre;
            }
        }
        return max_flow;
    }
    
    
    
    int main()
    {
        int i,u,v,cost,T,c,d,e,cas=0;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m);
            memset(gra,0,sizeof(gra));
            for(i=1;i<=m;i++){
                scanf("%d%d%d",&c,&d,&e);
                gra[c][d]+=e;
            }
            st=1;ed=n;
            printf("Case %d: %d
    ",++cas,Edmonds_Karp());
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/herumw/p/9464859.html
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