poj3259 Wormholes

Wormholes

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: 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.

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;
}