POJ No 3259 Wormholes Bellman-Ford 判断是否存在负图

题目：http://poj.org/problem?id=3259

题意：主要就是构造图， 然后判断，是否存在负图，可以回到原点

/*
2
3 3 1       //N, M, W

1 2 2
1 3 4
2 3 1

3 1 3       //虫洞 

3 2 1       //N, M, W

1 2 3
2 3 4

3 1 8

*/
#include <iostream>
#include <cstring>
using namespace std;

const int maxn = ( +  + ) *  + ;
const int INF =  + ;
int N, M, W;

//(from, to) 权值为cost
struct Edge {
    int from,
        to,    cost;
    Edge(int f = , int t = , int c = ) :
        from(f), to(t), cost(c) {}
};

//边
Edge es[maxn];

int d[maxn];       //最短距离
int V, E;          //顶点数，E边数 

bool find_negative_loop()
{
    memset(d, , sizeof(d));

    for (int i = ; i < V; i++)
    {
        for (int j = ; j < E; j++) {
            Edge e = es[j];
            if (d[e.to] > d[e.from] + e.cost) {
                d[e.to] = d[e.from] + e.cost;

                //如果第n次仍然更新了，则存在负圈
                if (i == V - ) return true;
            }
        }
    }
    return false;
}

void solve()
{
    int F;
    int from, to, cost;

    scanf("%d", &F);
    while (F--)
    {
        scanf("%d%d%d", &N, &M, &W); //顶点数，边数, 虫洞数
        V = N; E = ;                // E = M * 2 应该
        for (int i = ; i < M; ++i)
        {
            cin >> from >> to >> cost;
            --from; --to;
            //无向图 -- 去
            es[E].from = from; es[E].to = to;
            es[E].cost = cost; ++E;
            //回 -- 再来一次
            es[E].from = to; es[E].to = from;
            es[E].cost = cost; ++E;
        }

        for (int i = ; i < W; i++)
        {
            cin >> from >> to >> cost;
            --from; --to;
            es[E].from = from;
            es[E].to = to;
            //虫洞 - 回路
            es[E].cost = -cost;
            ++E;
        }
        if (find_negative_loop()) {
            printf("YES\n");
        } else {
            printf("NO\n");
        }
    }
}

int main()
{
    solve();

    return ;

}