POj-3259 Wormholes-Floyd判断负环

版权声明：欢迎转载 https://blog.csdn.net/suntengnb/article/details/80082814

问题描述：
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代码：

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]);//超时
                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)//超时
//        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));//初始化邻接矩阵
        for(int i = 1; i <= n; ++i)//初始化邻接矩阵
            f[i][i] = 0;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(f[a][b] > c)//取较小值
            {
                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;//直接覆盖
        }
        if(floyd() == true)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

解决方法：
正常的道路花费时间，虫洞穿越时间。题目要求农夫从某一农场出发，经过若干农场后再回到该农场，时光发生倒流。题目需要判断负环的存在性。我们求出各农场自回路的最短路径，若其为负，则负环存在。因为要求的是最短路，与虫洞同向的边没有存在的必要。
min函数稍微慢了点，会被卡。
注意在Floyd中，k是阶段，所以必须置于外层循环中，i和j是附加状态，所以应置于内层循环。