Wormholes
Time Limit: 2000MS | | Memory Limit: 65536K |
Total 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 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
#include<string.h>
#include<stdio.h>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
using 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;
}
