题目：

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

/*23 3 1       //N, M, W1 2 21 3 42 3 13 1 3       //虫洞 3 2 1       //N, M, W1 2 32 3 43 1 8*/#include 
     
      #include 
      
       using namespace std;const int maxn = (2500 + 200 + 16) * 2 + 24;const int INF = 10000 + 10;int N, M, W;//(from, to) 权值为cost struct Edge {    int from,        to,    cost;    Edge(int f = 0, int t = 0, int c = 0) :        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, 0, sizeof(d));        for (int i = 0; i < V; i++)     {        for (int j = 0; 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 - 1) 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 = 0;                // E = M * 2 应该         for (int i = 0; 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 = 0; 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 0;    }