poj 3259

Wormholes Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 48470 Accepted: 17868

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

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.

题解：在不同的农场有不同的距离，然后又虫洞可以穿越时空，问从某一地点出发，他是否可以再回到原点看到他自己，即是否可以穿越时空，那么在农场间距离为正值，虫洞间即为负值，对于含有负权值的图，不能用迪杰斯特拉算法，而且题目要求从不同的地点出发，即为多元路径，用弗洛伊德算法。最后计算map[i][i]是否为负值，如果有则可以穿越时空。

#include<stdio.h> #define INF 0x3f3f3f3f int distance[520][520]; int floya(int n) { int i, j, k; for(k = 1; k <= n; k++) for(i = 1; i <= n; i++) { for(j = 1; j <= n; j++) if(distance[i][j] > distance[i][k] + distance[k][j]) distance[i][j] = distance[i][k] + distance[k][j]; if(distance[i][i] < 0) return 1; } return 0; } int main() { int t, i, j, x, y, z, n, m, w; scanf("%d", &t); while(t--) { scanf("%d%d%d", &n, &m, &w); for(i = 1; i <= 519; i++) for(j = 1; j <= 519; j++) { if(i != j) distance[i][j] = INF; else distance[i][j] = 0; } for(i = 1; i <= m; i++) { scanf("%d%d%d", &x, &y, &z); if(distance[x][y] > z) { distance[x][y] = z; distance[y][x] = z; } } for(i = 1; i <= w; i++) { scanf("%d%d%d", &x, &y, &z); if(distance[x][y] > -z) distance[x][y] = -z; } int flag = floya(n); if(flag) printf("YES\n"); else printf("NO\n"); } }