就是给出一些点边关系， 有虫洞的就是负值。 注意普通道路是双向边。。一开始这里建错图了
之后处理用了spfa ，类似模板的一道题
题目：
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 26848 | Accepted: 9669 |
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
1 #include <iostream> 2 #include <vector> 3 #include <cstring> 4 #include <queue> 5 #include <cstdio> 6 using namespace std; 7 #define LEN 500+10 8 #define INF 10000*501 9 #define MP(x,y) make_pair(x,y) 10 typedef pair<int,int> pii; 11 12 vector<pii> G[LEN]; 13 int N,M,W; 14 int dis[LEN]; 15 void init() 16 { 17 cin>>N>>M>>W; 18 for(int i=1;i<=N;i++) 19 G[i].clear(); 20 21 int s,e,t; 22 for(int i=0;i<M;i++) 23 { 24 cin>>s>>e>>t; 25 G[s].push_back( MP(e,t)); 26 G[e].push_back( MP(s,t)); 27 } 28 for(int i=0;i<W;i++) 29 { 30 cin>>s>>e>>t; 31 G[s].push_back( MP(e,-t)); 32 } 33 } 34 35 bool spfa(int s) 36 { 37 queue<int> q; 38 int vis[LEN] = {0}, cnt[LEN] ={0}; 39 for(int i=1;i<=N;i++)dis[i] = INF; 40 41 dis[s]=0; 42 q.push(s); 43 vis[s] = 1; 44 cnt[s]++; 45 46 while(!q.empty()) 47 { 48 int v = q.front(); q.pop(); 49 for(int i=0;i<G[v].size();i++) 50 { 51 int x = G[v][i].first , d = G[v][i].second; 52 53 if( dis[x] > dis[v]+ d) 54 { 55 dis[x] = dis[v]+d; 56 if(!vis[x]) 57 { 58 q.push(x); 59 vis[x] = 1; 60 cnt[x] ++; 61 if( cnt[x]>N-1)return true; 62 } 63 } 64 } 65 vis[v]= 0; 66 } 67 return false; 68 } 69 int main() 70 { 71 int F; 72 cin>>F; 73 while(F--) 74 { 75 init(); 76 77 if( spfa(1) || dis[1]<0) 78 cout<<"YES"<<endl; 79 else 80 cout<<"NO"<<endl; 81 } 82 return 0; 83 }