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  • HDU 3549 Dinic

    Flow Problem

    Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 8946    Accepted Submission(s): 4205


    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
     
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     跟上一道题一样。
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstring>
    #include<vector>
    #include<queue>
    #include<algorithm>
    using namespace std;
    const int inf=0x3fffffff;
    const int maxn=207;
    struct Edge
    {
        int cap,flow;
    };
    int n,m,s,t;
    Edge edges[maxn][maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void init()
    {
              memset(edges,0,sizeof(edges));
    }
    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int>Q;
        Q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!Q.empty()){
            int x=Q.front();Q.pop();
            for(int i=1;i<=n;i++){
                Edge &e=edges[x][i];
                if(!vis[i]&&e.cap>e.flow){
                    vis[i]=1;
                    d[i]=d[x]+1;
                    Q.push(i);
                }
            }
        }
        return vis[t];
    }
    int DFS(int u,int cp)
    {
        int tmp=cp;
        int v,t;
        if(u==n)
            return cp;
        for(v=1;v<=n&&tmp;v++)
        {
            if(d[u]+1==d[v])
            {
                if(edges[u][v].cap>edges[u][v].flow)
                {
                    t=DFS(v,min(tmp,edges[u][v].cap-edges[u][v].flow));
                    edges[u][v].flow+=t;
                    edges[v][u].flow-=t;
                    tmp-=t;
                }
            }
        }
        return cp-tmp;
    }
    int Maxflow()
    {
        int flow=0;
        while(BFS()){
            memset(cur,0,sizeof(cur));
            flow+=DFS(s,inf);
        }
        return flow;
    }
    int main()
    {
      // freopen("in.txt","r",stdin);
       int tt,cnt=1;
       scanf("%d",&tt);
        int a,b, c;
        while(tt--){
            scanf("%d%d",&n,&m);
            init();
            for(int i=0;i<m;i++)
            {
                scanf("%d%d%d",&a,&b,&c);
                edges[a][b].cap+=c;
            }
            s=1;t=n;
            int ans=Maxflow();
            printf("Case %d: %d
    ",cnt++,ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/codeyuan/p/4262867.html
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