hdu3549Flow Problem【最大流】

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6817    Accepted Submission(s): 3178

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

 

Author

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

题意：最大流

分析：最大流模板

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

const int maxn = 20  << 1;
const int INF = 1000000000;

struct Edge
{
    int from, to, cap, flow;
};

struct Dinic
{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int>G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void ClearAll(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} );
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
        //printf("%din end
",m);
    }

    bool BFS()
    {
        memset(vis, 0, sizeof(vis) );
        queue<int> Q;
        Q.push(s);
        vis[s] = 1;
        d[s] = 0;
        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] = 1;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        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 s, int t) {
        this -> s = s; this -> t = t;
        int flow = 0;
        while(BFS()) {
            memset(cur, 0, sizeof(cur) );
            flow += DFS(s, INF);
        }
        return flow;
    }
};

Dinic g;

int main()
{
    int t;
    scanf("%d",&t);
    int n, m;
    int a, b, c;
    for(int kase = 1; kase <= t; kase++) {
        scanf("%d %d",&n, &m);
        g.ClearAll(maxn);
        for(int i = 0;i < m; i++ ) {
            scanf("%d %d %d",&a, &b, &c);
            g.AddEdge(a, b, c);
        }
        printf("Case %d: %d
",kase, g.Maxflow(1, n) );
    }
    return 0;
}