| 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
#include<stdio.h>
#include<string.h>
#define MAX 0x3f3f3f3f
struct path
{
int u , v , t ;
}pa[]; int d[] ;
int n , m , w ;
int s , e , t ;
int f ;
int cnt ; bool Bellman_ford ()
{
for (int i = ; i <= n ; i++)
d[i] = MAX ;
d[] = ;
bool flag ;
for (int i = ; i <= n ; i++) {// ' = ' 不能省
flag = ;
for (int j = ; j < cnt ; j++) {
if (d[pa[j].v] > d[pa[j].u] + pa[j].t) {
flag = ;
d[pa[j].v] = d[pa[j].u] + pa[j].t ;
}
}
if (flag)
return true ;
}
return false ;
} int main ()
{
//freopen ("a.txt" , "r" , stdin) ;
scanf ("%d" , &f) ;
while (f--) {
cnt = ;
scanf ("%d%d%d" , &n , &m , &w) ;
for (int i = ; i < m ; i++) {
scanf ("%d%d%d" , &s , &e , &t) ;
pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = t ;
cnt++ ;
pa[cnt].u = e , pa[cnt].v = s , pa[cnt].t = t ;
cnt++ ;
}
for (int i = ; i < w ; i++ , cnt++) {
scanf ("%d%d%d" , &s , &e , &t) ;
pa[cnt].u = s , pa[cnt].v = e , pa[cnt].t = -t ;
}
if (Bellman_ford())
puts ("NO") ;
else
puts ("YES") ;
}
return ;
}