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:
    grep命令
    Linux下tar.xz结尾的文件的解压方法
    const char*, char const*, char*const的区别
    "undefined reference to" 多种可能出现的问题解决方法
    Linux查找含有某字符串的所有文件
    Netbeans C++ unable to resolve identifier 无法解析标识符
    Linux 下编译C程序的全过程
    linux tar.gz zip 解压缩 压缩命令
    安装anaconda
    Mongodb数据迁移步骤
  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798633.html
Copyright © 2011-2022 走看看