Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 60400 | Accepted: 22564 |
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
题意:John的农场里N块地,M条路连接两块地(无向)没走过一条路,消耗w秒,V个虫洞(有向),走过虫洞,时间会倒退w秒。判段,是否可以以N块地中的一块地为起点,找到一条路,使得时间可以不断的倒退。
思路:spfa判断是否有负环!
#include "iostream" #include "queue" #include "vector" #include "cstring" using namespace std; const int Max=1e3+10; int N,M,V; int vis[Max],d[Max],cnt[Max];//分别记录是否已经入队,时间消耗,和入队次数 struct edge//记录目的地和话费,如果是虫洞的话,花费为负数 { int to,cost; }; vector<edge> G[Max]; bool spfa(int s) { vis[s]=1; d[s]=0; queue<int> que; que.push(s); while(!que.empty()){ int u=que.front(); que.pop(); vis[u]=0; for(int i=0;i<G[u].size();i++){ edge e=G[u][i]; if(d[e.to]>d[u]+e.cost){ d[e.to]=d[u]+e.cost; if(vis[e.to]) continue; que.push(e.to); vis[e.to]=1; if(++cnt[e.to]>N&&d[e.to]<0) return 1;//一个最多可以进队N次,如果大于N说明有环,如果时间为负数,说明是负环 } } } return 0; } int main() { ios::sync_with_stdio(false); int t; cin>>t; while(t--){ for(int i=0;i<Max;i++){G[i].clear(); vis[i]=0; d[i]=0xfffffff;cnt[i]=0;} int u,v,w; edge e; cin>>N>>M>>V; for(int i=0;i<M;i++){//普通道路 cin>>u>>v>>w; e.to=v,e.cost=w; G[u].push_back(e); e.to=u; G[v].push_back(e); } for(int i=0;i<V;i++){//虫洞 cin>>u>>v>>w; e.to=v; e.cost=-w; G[u].push_back(e); } int flag=0; for(int i=1;i<=N;i++) if(spfa(i)){flag=1;break;} if(flag) cout<<"YES"<<endl; else cout<<"NO"<<endl; } }