zoukankan      html  css  js  c++  java
  • 网络流--最大流--HDU 3549 Flow Problem

    题目链接

    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

    模板题,用来测试模板,我的模板都能过。

    #include<cstdio>
    #include<cstring>
    #include<queue>
    #define INF 1e9
    using namespace std;
    const int maxn=200+5;
    
    struct Edge
    {
        int from,to,cap,flow;
        Edge() {}
        Edge(int f,int t,int c,int flow):from(f),to(t),cap(c),flow(flow) {}
    };
    
    struct Dinic
    {
        int n,m,s,t;
        vector<Edge> edges;
        vector<int> G[maxn];
        bool vis[maxn];
        int cur[maxn];
        int d[maxn];
    
        void init(int n,int s,int t)
        {
            this->n=n, this->s=s, this->t=t;
            edges.clear();
            for(int i=1; i<=n; i++)
                G[i].clear();
        }
    
        void AddEdge(int from,int to,int cap)
        {
            edges.push_back(Edge(from,to,cap,0));
            edges.push_back(Edge(to,from,0,0));
            m = edges.size();
            G[from].push_back(m-2);
            G[to].push_back(m-1);
        }
    
        bool BFS()
        {
            memset(vis,0,sizeof(vis));
            queue<int> Q;
            d[s]=0;
            Q.push(s);
            vis[s]=true;
            while(!Q.empty())
            {
                int x=Q.front();
                Q.pop();
                for(int i=0; i<G[x].size(); i++)
                {
                    Edge& e=edges[G[x][i]];
                    if(!vis[e.to] && e.cap>e.flow)
                    {
                        vis[e.to]=true;
                        Q.push(e.to);
                        d[e.to]= 1+d[x];
                    }
                }
            }
            return vis[t];
        }
    
        int DFS(int x,int a)
        {
            if(x==t || a==0)
                return a;
            int flow=0,f;
            for(int& i=cur[x]; i<G[x].size(); i++)
            {
                Edge& e=edges[G[x][i]];
                if(d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow) ))>0 )
                {
                    e.flow+=f;
                    edges[G[x][i]^1].flow -=f;
                    flow+=f;
                    a-=f;
                    if(a==0)
                        break;
                }
            }
            return flow;
        }
    
        int Maxflow()
        {
            int flow=0;
            while(BFS())
            {
                memset(cur,0,sizeof(cur));
                flow += DFS(s,INF);
            }
            return flow;
        }
    } DC;
    
    int main()
    {
        int n,m,t;
        scanf("%d",&t);
        for(int i=1;i<=t;++i)
        {
            scanf("%d%d",&n,&m);
            DC.init(n,1,n);
            while(m--)
            {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                DC.AddEdge(u,v,w);
            }
            printf("Case %d: %d
    ",i,DC.Maxflow());
        }
        return 0;
    }
    
  • 相关阅读:
    IBM:linuxdeveloperweb
    xen虚拟化及工作原理
    (转) mysql 8小时空闲后连接超时的问题
    (转) 分布式缓存系统Memcached简介与实践
    ubuntu server 使用memcachetop监测memcache集群
    心安
    (转)Mysql导出表结构及表数据 mysqldump用法
    一个老工程师的心理话
    memcache设置set过期时间
    (转)memcached配置及命令
  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798633.html
Copyright © 2011-2022 走看看