POJ 3259 Wormholes
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 31992 | Accepted: 11614 |
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.
Source
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <queue> #include <map> #include <string> using namespace std; typedef long long LL; const int N = 1010; const int M = 20010; const int inf = 1e9+7; int n , m1 , m2 ; int eh[N] , et[M] , nxt[M] ,ew[M] , tot ; bool inq[N] , vis[N] ; int dis[N] , in_cnt[N]; void init(){ tot = 0 ; memset( eh , -1 , sizeof eh ); memset( vis , false , sizeof vis ); } void addedge( int u , int v ,int w ) { et[tot] = v , ew[tot] = w , nxt[tot] = eh[u] , eh[u] = tot++ ; } bool spfa( int s ) { queue<int>que; memset( inq , false, sizeof inq ); memset( in_cnt , 0, sizeof in_cnt ); for( int i = 0 ; i <= n ; ++i ) dis[i] = inf ; que.push(s) , inq[s] = true , dis[s] = 0 , in_cnt[s]++ ; while( !que.empty() ) { int u = que.front() ; que.pop(); inq[u] = false; vis[u] = true; for( int i = eh[u] ; ~i ; i = nxt[i] ) { int v = et[i] , w = ew[i] ; if( dis[u] + w < dis[v] ){ dis[v] = dis[u] + w ; if( !inq[v] ) { inq[v] = true; in_cnt[v]++; if( in_cnt[v] >= n ) return false; que.push(v); } } } } return true; } void run() { int u, v, w ; init(); scanf("%d%d%d",&n,&m1,&m2); while( m1-- ) { scanf("%d%d%d",&u,&v,&w); addedge( u , v , w ); addedge( v , u , w ); } while( m2-- ) { scanf("%d%d%d",&u,&v,&w); addedge( u , v , -w ); } for( int i =1 ; i <= n ; ++i ) if(!vis[i]) { if( !spfa(i) ) { puts("YES"); return ; } } puts("NO"); } int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL int _ , cas = 1 ; cin >> _ ; while( _-- ) run(); }