POJ 3259 Wormholes

Description

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

Sample Output

NO
YES

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

#define inf 0x3f3f3f3f
int mp[510][510];
int N, M, W;

int floyd() {
    for(int k = 1; k <= N; k ++) {
        for(int i = 1; i <= N; i ++) {
            for(int j = 1; j <= N; j ++) {
                if(mp[i][k] + mp[k][j] < mp[i][j])
                    mp[i][j] = mp[i][k] + mp[k][j];
            }
            if(mp[i][i] < 0)
                return 1;
        }
    }
    return 0;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        int a, b, c;
        scanf("%d%d%d", &N, &M, &W);
        memset(mp, inf, sizeof(mp));
        
        for(int i = 1; i <= N; i ++)
            mp[i][i] = 0;
        
        for(int i = 1; i <= M; i ++) {
            scanf("%d%d%d", &a, &b, &c);
            if(c < mp[a][b])
                mp[a][b] = mp[b][a] = c;
        }
        
        for(int i = 1; i <= W; i ++) {
            scanf("%d%d%d", &a, &b, &c);
            mp[a][b] = -c;
        }
        
        int ans = floyd();
        if(!ans)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}