F

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
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

题解：

Bellman-Ford判负环。
很好玩的虫洞问题。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 550;int dist[maxn];struct Edge{    int u;    int v;    int cost;    Edge(int _u,int _v,int _cost):u(_u),v(_v),cost(_cost){}};vector<Edge> E;bool BellmanFord(int start,int n){   memset(dist,0x3f,sizeof(dist));   dist[start]=0;   for(int i=1;i<n;i++)   {       bool flag = false;       for(int j=0;j<E.size();j++)       {           int u = E[j].u;           int v = E[j].v;           int cost = E[j].cost;           if(dist[v]>dist[u]+cost)           {               dist[v] = dist[u]+cost;               flag = true;           }       }       if(!flag) return true;   }   for(int j=0;j<E.size();j++)   {       if(dist[E[j].v]>dist[E[j].u]+E[j].cost)       {           return false;       }   }   return true;}int main(){    int T;    scanf("%d",&T);    int N,M,W;    while(T--)    {       scanf("%d%d%d",&N,&M,&W);       int u,v,t;       for(int i=1;i<=M;i++)       {        scanf("%d%d%d",&u,&v,&t);        E.push_back(Edge(u,v,t));        E.push_back(Edge(v,u,t));       }       for(int i=1;i<=W;i++)       {           scanf("%d%d%d",&u,&v,&t);           E.push_back(Edge(u,v,-t));       }       bool flag = BellmanFord(1,N);       if(flag) cout<<"NO"<<endl;       else cout<<"YES"<<endl;          E.clear();    }    return 0;}