POJ 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 32555 Accepted: 11831

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

题意：约翰希望从他的某块地出发并且回到这块地，使得他回来的时间早于出发的时间，两块地之间有路径或虫洞，路径是双向的，虫洞是单向的，虫洞可以让时光倒流。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int INF=0x3f3f3f3f;int sum;int n,m,ss;int dis[510];struct Edge{int u;int v;int w;}edge[5300];int BF(){for(int i=0;i<=n;i++)//手残到这种程度也真的是醉了dis[i]=INF;dis[0]=0;for(int i=1;i<n;i++){int flag=0;for(int j=0;j<sum;j++){if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w){dis[edge[j].v]=dis[edge[j].u]+edge[j].w;flag=1;}}if(!flag)//如果某次迭代中没有任何一个dis值改变，说明该遍迭代中的所有边均未被松弛，至此后，所有的边都不需再松弛，可直接退出迭代break;}for(int i=0;i<sum;i++){if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)return 1;}return 0;}int main(){int T;//int n,m,w;int s,e,t;scanf("%d",&T);while(T--){sum=0;scanf("%d%d%d",&n,&m,&ss);for(int i=0;i<m;i++){scanf("%d%d%d",&s,&e,&t);//路径是双向的edge[sum].u=s;edge[sum].v=e;edge[sum].w=t;sum++;edge[sum].u=e;edge[sum].v=s;edge[sum].w=t;sum++;}for(int i=0;i<ss;i++){scanf("%d%d%d",&s,&e,&t);//虫洞是单向的edge[sum].u=s;edge[sum].v=e;edge[sum].w=-t;//虫洞的作用是时间倒流，因此要把权值设成负值sum++;}if(BF())printf("YES\n");elseprintf("NO\n");}return 0;}