Flow 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
一个裸的最大流问题。
基本思路就是不断找到通路，再把通路上的流量减掉，一直到没有通路为止，再减去流量的同时，还可以建立反向边，再把流量加上去，这样就可以给了一个回溯的机会。
紫书上直接把它封装了- -

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 2005;


struct Edge{
    int form, to, cap, flow;
    Edge(int u, int v, int c, int f) :form(u), to(v), cap(c), flow(f){}
};
struct EdmondKarp{
    int n, m;
    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));
        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;
            Q.push(s);
            a[s] = inf;
            while (Q.size())
            {
                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 u = t; u != s; u = edges[p[u]].form)
            {
                edges[p[u]].flow += a[t];
                edges[p[u] ^ 1].flow -= a[t];
            }
            flow += a[t];
        }
        return flow;
    }
}ans;
int main()
{
    int i, j, m, n, t, p = 1;
    cin >> t;
    while (t--)
    {
        cin >> n >> m;
        ans.init(n);
        for (i = 1; i <= m; i++)
        {
            int x, y, z;
            cin >> x >> y >> z;
            ans.AddEdge(x - 1, y - 1, z);
        }
        cout << "Case " << p++ << ": ";
        cout << ans.Maxflow(0, n - 1) << endl;
    }
    return 0;
}