最短路(Bellman_Ford) POJ 3259 Wormholes

题目传送门

/*
    题意：一张有双方向连通和单方向连通的图，单方向的是负权值，问是否能回到过去(权值和为负)
    Bellman_Ford：循环n-1次松弛操作，再判断是否存在负权回路(因为如果有会一直减下去)
    注意：双方向连通要把边起点终点互换后的边加上
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 6000 + 10;
const int INF = 0x3f3f3f3f;
int d[MAXN];
struct NODE
{
    int from, to, cost;
}node[MAXN];

bool Bellman_Ford(int s, int n, int m)
{
    int x, y;
    d[s] = 0;
    for (int k=1; k<=n-1; ++k)
    {
        for (int i=1; i<=m; ++i)
        {
            NODE e = node[i];
            d[e.to] = min (d[e.to], d[e.from] + e.cost);
        }
    }

    for (int i=1; i<=m; ++i)
    {
        NODE e = node[i];
        if (d[e.to] > d[e.from] + e.cost)    return false;
    }

    return true;
}

void work(int n, int m)
{
    bool flag = false;
    flag = Bellman_Ford (1, n, m);

    (!flag) ? puts ("YES") : puts ("NO");
}

int main(void)        //POJ 3259 Wormholes
{
    //freopen ("B.in", "r", stdin);

    int N, M, W;
    int t;
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d%d%d", &N, &M, &W);
        for (int i=1; i<=N; ++i)    d[i] = INF;

        int x, y, z;
        for (int i=1; i<=2*M; ++i)
        {
            scanf ("%d%d%d", &x, &y, &z);
            node[i].from = x;    node[i].to = y;        node[i].cost = z;
            node[++i].from = y;        node[i].to = x;        node[i].cost = z;
        }

        for (int i=2*M+1; i<=2*M+W; ++i)
        {
            scanf ("%d%d%d", &x, &y, &z);
            node[i].from = x;    node[i].to = y;        node[i].cost = -z;
        }

        work (N, 2*M+W);
    }

    return 0;
}