POJ 3059 Wormholes

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.

Input
Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Lines 1.. F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
Sample Input
2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8
Sample Output
NO
YES
Hint
For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
题意：
这个人在F个农场上做实验，每个农场有N块田地，田地之间有M条路，W个虫洞。接着输M行，每行三个数据，分别表示田地编号、田地编号、两田地走路需要的时间。又输入W行，每行三个数据，分别表示田地编号、田地编号、通过虫洞回溯的时间。判断这个人能否在某块田地上出发，经过一系列路和虫洞后，在自己出发之前赶回来。
需要注意，两块田地之间可以有多条路，在赋值时要选择最短的一条赋值。

Floyd-Warshall算法，（特别容易超时）

#include<iostream>#include<vector>#include<algorithm>#include<cstdlib>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<cstring>#include<string.h>#include<map>#include<set>using namespace std;#define N 1000+5#define NN 500000+5#define INF 0x3f3f3f3f/*****************************************************/int d[NN];int cost[N][N];int n, m, mm;struct node{    int u, v;    int w;};node s[NN];bool find(){    int j=1;    for (int i = 1; i <= n; i++){        for ( j = 1; j <= n; j++){            for (int k = 1; k <= n; k++){                int t = cost[j][i] + cost[i][k];                if (t < cost[j][k])                    cost[j][k] = t;            }        }        if (cost[i][i] < 0)return true;    }    return false;}int main(){    int t;    cin >> t;    while (t--){        cin >> n >> m >> mm;        int edge = 0;        memset(cost, 0x3f, sizeof(cost));        for (int i = 0; i < m; i++){            int u, v; int w;            scanf("%d%d%d", &u, &v, &w);            if (w < cost[u][v])     //在这里选择多条路中的最小路                cost[u][v] = cost[v][u]=w;      //无权边        }        for (int i = m; i < m + mm; i++){            int u, v; int w;            scanf("%d%d%d", &u, &v, &w);            cost[u][v] = -w;        }        if (find())  cout << "YES" << endl;        else         cout << "NO" << endl;    }}