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.
#include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; vector< pair<int,int> > g[20000]; int d[20000],vis[20000],cnt[20000]; int ttt,n,m,w; int spfa() { memset(vis,0,sizeof(vis)); memset(cnt,0,sizeof(cnt)); d[1]=0; queue<int> q; q.push(1); cnt[1]++; vis[1]=1; while(!q.empty()) { int x=q.front(); vis[x]=0; q.pop(); int sz=g[x].size(); for(int i=0; i<sz; i++) { int y=g[x][i].first; int w=g[x][i].second; if(d[x]+w<d[y]) { d[y]=d[x]+w; if(!vis[y]) { q.push(y); vis[y]=1; cnt[y]++; if(cnt[y]>n) { return 1; } } } } } return 0; } int main() { scanf("%d",&ttt); while(ttt--) { scanf("%d%d%d",&n,&m,&w); for(int i=1;i<=n;i++) { g[i].clear(); } for(int i=1; i<=m; i++) { int f,t,c; scanf("%d%d%d",&f,&t,&c); g[f].push_back(make_pair(t,c)); g[t].push_back(make_pair(f,c)); } for(int i=1; i<=w; i++) { int f,t,c; scanf("%d%d%d",&f,&t,&c); g[f].push_back(make_pair(t,-c)); } for(int i=1; i<=n; i++) { d[i]=inf; } int ans=spfa(); if(ans) { printf("YES\n"); } else { printf("NO\n"); } } }
