HDOJ3549 最大流裸题，贴模板程序

最新推荐文章于 2018-11-13 22:18:59 发布

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本文详细介绍了一种经典的网络流问题——最大流问题，并提供了一个使用EK算法解决该问题的具体实例。通过邻接矩阵的方式存储图数据，实现了一个求解从源点到汇点的最大流的程序。

Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 17088 Accepted Submission(s): 8039

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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刚自学了EK算法，因为数据不大，就用了邻接矩阵存储

#include <iostream>
#include <queue>
#include <ctime>
#include <cmath>
#include <cstring>
using namespace std;

const int maxn = 20;
const int inf = 1<<30;
int flow[maxn],c[maxn][maxn],pre[maxn],n,m;
queue<int>q;

int bfs() {
    memset(flow,0,sizeof(flow));
    memset(pre,-1,sizeof(pre));
    pre[1] = 0;
    flow[1] = inf;
    while (!q.empty()) q.pop();
    q.push(1);
    int tmp;

    while (!q.empty()) {
        tmp = q.front();
        q.pop();
        if (tmp==n) break;

        for (int i=1; i<=n; i++) {
            if (pre[i]==-1 && i!=1 && c[tmp][i]) {
                flow[i] = min(flow[tmp],c[tmp][i]);
                pre[i] = tmp;
                q.push(i);
            }
        }
    }

    if (pre[n]==-1) return -1;
    return flow[n];
}

int EK(){
    int max_flow = 0;
    int increase,tmp,k;
    while ((increase=bfs())!=-1) {
        max_flow += increase;
        tmp = n;
        while (tmp!=1){
            k=pre[tmp];
            c[k][tmp] -= increase;
            c[tmp][k] += increase;
            tmp = k;
        }
    }
    return max_flow;
}

int main(){
    std::ios::sync_with_stdio(false);
    int t,s,e,tmp;
    cin >> t;
    for (int j=1; j<=t; j++) {
         cin >> n >> m;
         memset(c,0,sizeof(c));
         for (int i=1; i<=m; i++) {
             cin >> s >> e >> tmp;
             if (s==e) continue;
             c[s][e] += tmp;
         }
         cout << "Case " << j << ": " << EK() << endl;
    }
    return 0;
}