HDU3549Dinic

/*
HDU3549网络最大流Dinic
题意：给定一个图，找出加权有向图的最大流
输入的第一行包含一个整数T，表示测试用例的数目。
对于每一个测试案例，第一行包含两个整数n和m，表示图中顶点和边数。
接下来的m行，每行包含三个整数x，y和z，表示从x到y的一条边的容量是z
对于每一组测试样例，输出从1到n的最大流量。
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
*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#define maxn 10010
#define INF 0x3f3f3f3f
using namespace std;

struct Edge
{
    int from,to,cap,flow;
};
bool operator < (const Edge& a, const Edge& b)
{
    return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic
{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void Init(int n)
    {
        this->n = 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);
    }
    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int>Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;
        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 num = 0;
    int T,n,m,x,y,z;
    scanf("%d",&T);
    while(T --)
    {
        scanf("%d%d",&n,&m);
        g.Init(n);
        for(int i = 0; i < m; i ++)
        {
            scanf("%d%d%d",&x,&y,&z);
            g.Addedge(x - 1,y - 1,z);
        }
        int flow = g.Maxflow(0,n-1);
        printf("Case %d: %d\n",++num,flow);
    }
    return 0;
}