HDU-3549-Flow Problem

#include <iostream>
#include <memory.h>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <queue>
#include <string>

using namespace std;

typedef long long LL;

const int MAXM = 1000 + 10;
const int MAXN = 20;
const int INF = 1e9 + 10;

int n, m;

struct Edge
{
    int _from, _to, _cap, _flow;
    Edge(int from, int to, int cap, int flow):_from(from), _to(to), _cap(cap), _flow(flow){};
};

vector<Edge> edges;//记录每条边
vector<int> G[MAXN];//记录图,每个节点的边
int a[MAXN], p[MAXN];
//a数组记录每次找增广路时候每个节点对应能加的流
//p数组记录流经的点对应的线路的编号
void Init()
{
    edges.clear();
    for (int i = 0;i < n;i++)
        G[i].clear();
}

int Solve()
{
    int flow = 0;//总流量
    while (1)
    {
        memset(a, 0, sizeof(a));//每次找增广路初始化
        queue<int> que;//每次bfs的队列
        que.push(1);
        a[1] = INF;
        while (!que.empty())
        {
            int x = que.front();
            que.pop();
            for (int i = 0;i < G[x].size();i++)
            {
                Edge & e = edges[G[x][i]];
                if (!a[e._to] && e._cap > e._flow)
                {
                    //a为0表示没有流过这个点同时流量小于容量
                    p[e._to] = G[x][i];
                    a[e._to] = min(a[x], e._cap - e._flow);
                    //能流的流量是线路最大流量和上一个节点的值中的较小值
                    que.push(e._to);//新节点进队
                }
            }
            if (a[n] > 0)//如果流到了终点,break
                break;
        }
        if (a[n] == 0)//如果终点流不到，表示增广路找不到，
            break;
        for (int u = n;u != 1;u = edges[p[u]]._from)
        {
            edges[p[u]]._flow += a[n];
            edges[p[u] ^ 1]._flow -= a[n];//表示与正向对应的反向的边
        }
        flow += a[n];
    }
    return flow;
}

int main()
{
    int t;
    int l, r, c, cnt = 0;
    scanf("%d", &t);
    while (t--)
    {

        scanf("%d%d", &n, &m);
        Init();
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d%d", &l, &r, &c);
            edges.push_back(Edge(l, r, c, 0));
            edges.push_back(Edge(r, l, 0, 0));//增加反向边 保证能找到最大流
            G[l].push_back(edges.size() - 2);
            G[r].push_back(edges.size() - 1);
        }
        cout << "Case " << ++cnt << ": ";
        cout << Solve() << endl;
    }

    return 0;
}