Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
|---|---|---|
| Total Submissions: 34833 | Accepted: 12724 |
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: 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
//题目大意是此人能不能经过进虫洞时光倒流，就是裸的贝尔曼判环。
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m,w,z;
struct node
{
int a,b,c;
}q[100010];
int dis[100010];
void creat(int s,int e,int t)
{
q[z].a=s;
q[z].b=e;
q[z++].c=t;
}
int bf()
{
int i,j;
//memset(vis,0,sizeof(vis));//注意贝尔曼中不需要标记数组
for(i=1;i<=n;i++)
{
dis[i]=0x3f3f3f3f;
}
dis[1]=0;
//vis[1]=1;
for(i=0;i<n-1;i++)
{
for(j=0;j<z;j++)
{
if(dis[ q[j].b ]>dis[q[j].a ]+q[j].c)//弄清楚谁大于谁
{
dis[q[j].b ]=dis[ q[j].a ]+q[j].c;
}
}
}
for(i=0;i<z;i++)
{
if(dis[q[i].b ]>dis[q[i].a ]+q[i].c)
{
return 1;
}
}
return 0;
}
int main()
{
int s,e,t,i,j,k;
cin>>k;
while(k--)
{
z=0;
cin>>n>>m>>w;
for(i=0;i<m;i++)
{
cin>>s>>e>>t;
creat(s,e,t);
creat(e,s,t);
}
for(i=0;i<w;i++)
{
cin>>s>>e>>t;
creat(s,e,-t);
}
int p=bf();
if(p)
puts("YES");
else
puts("NO");
}
return 0;
}
