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 :) .文章来源地址:https://www.yii666.com/article/756145.html
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文章地址https://www.yii666.com/article/756145.html网址:yii666.com
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.
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std; vector< pair<int,int> > g[];
int d[],vis[],cnt[];
int ttt,n,m,w; int spfa()
{
memset(vis,,sizeof(vis));
memset(cnt,,sizeof(cnt));
d[]=;
queue<int> q;
q.push();
cnt[]++;
vis[]=;
while(!q.empty())
{
int x=q.front();
vis[x]=;
q.pop();
int sz=g[x].size();
for(int i=; i<sz; i++)
{
int y=g[x][i].first;
int w=g[x][i].second;
if(d[x]+w<d[y])
{
d[y]=d[x]+w;
if(!vis[y])
{
q.push(y);
vis[y]=;
cnt[y]++;
if(cnt[y]>n)
{
return ;
}
}
}
}
}
return ;
} int main()
{
scanf("%d",&ttt);
while(ttt--)
{
scanf("%d%d%d",&n,&m,&w);
for(int i=;i<=n;i++)
{
g[i].clear();
}
for(int i=; i<=m; i++)
{
int f,t,c;
scanf("%d%d%d",&f,&t,&c);
g[f].push_back(make_pair(t,c));
g[t].push_back(make_pair(f,c));
}
for(int i=; i<=w; i++)
{
int f,t,c;
scanf("%d%d%d",&f,&t,&c);
g[f].push_back(make_pair(t,-c));
}
for(int i=; i<=n; i++)
{
d[i]=inf;
}
int ans=spfa();
if(ans)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
}