POJ_3259(Bellman-ford)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 23486 Accepted: 8398

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.

#include "stdio.h"const int MaxLen=999999;const int MaxV = 1000;const int MaxE=5000;int dist[MaxV];int edge_num,point_num;struct edge{int u;int v;int weight;}edge[MaxE];bool bellman_ford(){int i,j;dist[1]=0;for(i=2;i<=point_num;i++)dist[i]=MaxLen;for(i=1;i<point_num;i++)for(j=1;j<=edge_num;j++)if(dist[edge[j].u]+edge[j].weight<dist[edge[j].v])dist[edge[j].v]=dist[edge[j].u]+edge[j].weight;for(j=1;j<=edge_num;j++)if(dist[edge[j].u]+edge[j].weight<dist[edge[j].v])return false;return true;}int main(int argc, char* argv[]){int farm,i,j,s,e,length,n,m,w,f,k=0;//FILE *fp = freopen("data.txt","r",stdin);scanf("%d",&f);for(farm=1;farm<=f;farm++){scanf("%d%d%d",&n,&m,&w);k=0;for(i= 1;i<=m;i++){scanf("%d%d%d",&s,&e,&length);k++;edge[k].u = s;edge[k].v = e;edge[k].weight = length;k++;edge[k].u=e;edge[k].v=s;edge[k].weight=length;}for(i= 1;i<=w;i++){scanf("%d%d%d",&s,&e,&length);k++;edge[k].u = s;edge[k].v = e;edge[k].weight = -length;}point_num = n;edge_num=k;if(!bellman_ford())                        printf("YES\n");                  else  printf("NO\n"); }return 0;}