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 :) .网址:yii666.com
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.文章地址https://www.yii666.com/article/756142.html
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文章来源地址:https://www.yii666.com/article/756142.html
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.
#include<iostream>
#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;
const int maxn=;
int T,x,y,z,n,m,k,u,v,w,cnt[maxn],dis[maxn];vector<int> v1[maxn],v2[maxn];bool vis[maxn];queue<int> que;
bool spfa(int s){
while(!que.empty())que.pop();
memset(vis,false,sizeof(vis));
que.push(s),vis[s]=true,cnt[s]++,dis[s]=;//注意：自己到本身的时间为dis[s]=0，同时累计每个顶点的入队次数，用于判负环
while(!que.empty()){
u=que.front(),que.pop(),vis[u]=false;
for(size_t j=;j<v1[u].size();++j){
v=v1[u][j],w=v2[u][j];
if(dis[u]+w<dis[v]){//松弛
dis[v]=dis[u]+w;
if(!vis[v]){
que.push(v),vis[v]=true;
if(++cnt[v]>n)return true;//如果第n+1次仍然更新，则存在负圈
}
}
}
}
return false;
}
int main(){
while(~scanf("%d",&T)){
while(T--){
scanf("%d%d%d",&n,&m,&k);memset(cnt,,sizeof(cnt));
for(int i=;i<=n;++i)v1[i].clear(),v2[i].clear();
for(int i=;i<=n;++i)dis[i]=2e9;
while(m--){
scanf("%d%d%d",&x,&y,&z);//建双向边
v1[x].push_back(y),v1[y].push_back(x);
v2[x].push_back(z),v2[y].push_back(z);
}
while(k--){
scanf("%d%d%d",&x,&y,&z);//建单向边，权值为负数，表示时间倒流
v1[x].push_back(y);
v2[x].push_back(-z);
}
if(spfa())puts("YES");
else puts("NO");
}
}
return ;
}