网络流--最大流--HDU 3549 Flow Problem

风骨散人Chiam

发布于 2020-10-28 11:32:58

2850

发布于 2020-10-28 11:32:58

文章被收录于专栏：CSDN旧文

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\n",i,DC.Maxflow());
    }
    return 0;
}

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原始发表：2019/11/02 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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