HDU 3549 Flow Problem-优快云博客

本文介绍了一个典型的最大流问题，通过增广路算法解决加权有向图中的最大流问题。文章提供了一段完整的C++代码实现，并详细解释了输入输出格式及样例。

Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 18435 Accepted Submission(s): 8676

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

Sample Output

Case 1: 1
Case 2: 2

Author

HyperHexagon

Source

HyperHexagon’s Summer Gift (Original tasks)

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题意

n个点，m条边，求最大流。

解题思路

增广路算法模板题。

代码

#include<stdio.h>
#include<iostream>
#include<vector>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 1005
#define inf 0x3f3f3f3f

int n,m;
struct Edge
{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f) {}
};
struct EdmondsKarp
{
    vector<Edge> edges;
    vector<int> G[maxn];
    int a[maxn];
    int p[maxn];

    void init(int n)
    {
        for(int i=0; i<n; i++)
            G[i].clear();
        edges.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));
        int m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    int Maxflow(int s,int t)
    {
        int flow=0;
        for(;;)
        {
            memset(a,0,sizeof(a));
            queue<int> q;
            a[s]=inf;
            q.push(s);
            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(!a[e.to]&&e.cap>e.flow)
                    {
                        p[e.to]=G[x][i];
                        a[e.to]=min(a[x],e.cap-e.flow);
                        q.push(e.to);
                    }
                }
                if(a[t]) break;
            }
            if(!a[t]) break;
            for(int i=t; i!=s; i=edges[p[i]].from)
            {
                edges[p[i]].flow+=a[t];
                edges[p[i]^1].flow-=a[t];
            }
            flow+=a[t];
        }
        return flow;
    }
};
int main()
{
//    freopen("in.txt","r",stdin);
    EdmondsKarp E;
    int t;
    cin>>t;
    for(int k=1; k<=t; k++)
    {
        printf("Case %d: ",k);
        scanf("%d%d",&n,&m);
        E.init(n);
        int a,b,c;
        for(int i=0; i<m; i++)
        {
            cin>>a>>b>>c;
            E.addedge(a,b,c);
        }
        printf("%d\n",E.Maxflow(1,n));
    }
    return 0;
}