問題描述：
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
AC程式碼：

int f[510][510];
int n,m,w;
inline bool floyd()
{
    for(int k = 1; k <= n; ++k)
    {
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= n; ++j)
            {
                //f[i][j] = min(f[i][j],f[i][k] + f[k][j]);//超時
                int d = f[i][k] + f[k][j];
                if
 
(f[i][j] > d)
                    f[i][j] = d;
            }
            if(f[i][i] < 0)
                return true;
        }
    }
//    for(int i = 1; i <= n; ++i)//超時
//        if(f[i][i] < 0)
//            return true;
    return false;
}
int main()
{
    int t;
    cin >> t;
    while
 
(t--)
    {
        int a,b,c;
        scanf("%d%d%d",&n,&m,&w);
        memset(f,0x3F,sizeof(f));//初始化鄰接矩陣
        for(int i = 1; i <= n; ++i)//初始化鄰接矩陣
            f[i][i] = 0;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(f[a][b] > c)//取較小值
            {
                f[a][b] = c;
                f[b][a] = c;
            }
        }
        for(int i = 1; i <= w; ++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            f[a][b] = -c;//直接覆蓋
        }
        if(floyd() == true)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

解決方法：
正常的道路花費時間，蟲洞穿越時間。題目要求農夫從某一農場出發，經過若干農場後再回到該農場，時光發生倒流。題目需要判斷負環的存在性。我們求出各農場自迴路的最短路徑，若其為負，則負環存在。因為要求的是最短路，與蟲洞同向的邊沒有存在的必要。
min函式稍微慢了點，會被卡。
注意在Floyd中，k是階段，所以必須置於外層迴圈中，i和j是附加狀態，所以應置於內層迴圈。