Wormholes 3259 （最短路+判断负环是否存在）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 36832 Accepted: 13494

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 toF (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 toE 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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

#include<string.h>#include<stdio.h>#include<queue>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;int n,m,t;struct Edge{int from,to,val,next;}edge[5100];int head[600],edgenum;void add(int u,int v,int w){Edge E={u,v,w,head[u]};edge[edgenum]=E;head[u]=edgenum++;}int dis[600],vis[600],used[600];void SPFA(int x){queue<int>q;memset(vis,0,sizeof(vis));memset(dis,INF,sizeof(dis));memset(used,0,sizeof(used));q.push(x);dis[x]=0;vis[x]=1;used[x]++;while(!q.empty()){int u=q.front();q.pop();vis[u]=0;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(dis[v]>dis[u]+edge[i].val){dis[v]=dis[u]+edge[i].val;if(!vis[v]){vis[v]=1;used[v]++;if(used[v]>n){printf("YES\n");return ;}q.push(v);}}}}printf("NO\n");}int main(){int T,a,b,c;scanf("%d",&T);while(T--){memset(head,-1,sizeof(head));edgenum=0;scanf("%d%d%d",&n,&m,&t);while(m--){scanf("%d%d%d",&a,&b,&c);add(a,b,c);add(b,a,c);}while(t--){scanf("%d%d%d",&a,&b,&c);add(a,b,-c);}SPFA(1);}return 0;}