| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 33008 | Accepted: 12011 |
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 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
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 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.
Source
1 #include<stdio.h> 2 #include<string.h> 3 #define MAX 0x3f3f3f3f 4 struct path 5 { 6 int u , v , t ; 7 }pa[6000]; 8 9 int d[6000] ; 10 int n , m , w ; 11 int s , e , t ; 12 int f ; 13 int cnt ; 14 15 bool Bellman_ford () 16 { 17 for (int i = 1 ; i <= n ; i++) 18 d[i] = MAX ; 19 d[1] = 0 ; 20 bool flag ; 21 for (int i = 1 ; i <= n ; i++) {// ' = ' 不能省 22 flag = 1 ; 23 for (int j = 0 ; j < cnt ; j++) { 24 if (d[pa[j].v] > d[pa[j].u] + pa[j].t) { 25 flag = 0 ; 26 d[pa[j].v] = d[pa[j].u] + pa[j].t ; 27 } 28 } 29 if (flag) 30 return true ; 31 } 32 return false ; 33 } 34 35 int main () 36 { 37 //freopen ("a.txt" , "r" , stdin) ; 38 scanf ("%d" , &f) ; 39 while (f--) { 40 cnt = 0 ; 41 scanf ("%d%d%d" , &n , &m , &w) ; 42 for (int i = 0 ; i < m ; i++) { 43 scanf ("%d%d%d" , &s , &e , &t) ; 44 pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = t ; 45 cnt++ ; 46 pa[cnt].u = e , pa[cnt].v = s , pa[cnt].t = t ; 47 cnt++ ; 48 } 49 for (int i = 0 ; i < w ; i++ , cnt++) { 50 scanf ("%d%d%d" , &s , &e , &t) ; 51 pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = -t ; 52 } 53 if (Bellman_ford()) 54 puts ("NO") ; 55 else 56 puts ("YES") ; 57 } 58 return 0 ; 59 }