Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 50389 | Accepted: 18606 |
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.
本质就是求是否存在负圈
//
// Created by Admin on 2017/5/2.
//
#include <cstdio>
#include <cstring>
struct edge{
int from,to,cost;
}ed[6000];
int n,m,w,d[510];
int shortpath(int s){
memset(d,0, sizeof(d));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < s; ++j) {
edge e=ed[j];
if(d[e.to]>d[e.from]+e.cost){
d[e.to]=d[e.from]+e.cost;
//如果第n次任然更新了,则存在负圈
if(i==n-1)return 1;
}
}
}
return 0;
}
int main(){
int f;
scanf("%d",&f);
while (f--){
scanf("%d%d%d",&n,&m,&w);
int cnt=0,a,b,c;;
for (int i = 0; i < m; ++i) { //双向边
scanf("%d%d%d",&a,&b,&c);
ed[cnt].from=a;
ed[cnt].to=b;
ed[cnt++].cost=c;
ed[cnt].from=b;
ed[cnt].to=a;
ed[cnt++].cost=c;
}
for (int j = 0; j < w; ++j) { //单向负边
scanf("%d%d%d",&a,&b,&c);
ed[cnt].from=a;
ed[cnt].to=b;
ed[cnt++].cost=-c;
}
if(shortpath(cnt))printf("YES\n");
else printf("NO\n");
}
return 0;
}