poj 3259 Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 20003 | Accepted: 7088 |
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 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
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.
Source
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。
存在负权回路输出YES, 否则NO.
#include <string.h>
#include <vector>
#include <queue>
#define INF 1000000000
using namespace std;
struct node
{
int to;
int w;
}tedge;
vector<node>edge[600];
int vis[600],cnt[600];
int dis[600];
int n,m,t;
void add( int a,int b,int c)
{
tedge.to=b;
tedge.w=c;
edge[a].push_back(tedge);
}
int spfa( int s)
{
int i;
queue<int>q;
memset(vis,0,sizeof(vis));
memset(cnt,0,sizeof(cnt));
dis[s]=0;
vis[s]=1;
q.push(s);
while(!q.empty( ))
{
int u=q.front( );
q.pop( );
vis[u]=0;
for( i=0;i<edge[u].size( );i++)
{
if(dis[edge[u][i].to]-edge[u][i].w>dis[u])
{
dis[edge[u][i].to]=edge[u][i].w+dis[u];
if(!vis[edge[u][i].to])
{
q.push(edge[u][i].to);
vis[edge[u][i].to]=1;
if(++cnt[edge[u][i].to]>n)
return -1;
}
}
}
}
return 1;
}
int main( )
{
int s,e,tt,i;
int test;
scanf("%d",&test);
while(test--)
{
scanf("%d%d%d",&n,&m,&t);
for( i=1;i<=n;i++)
dis[i]=INF;
for( i=1;i<=n;i++)
edge[i].clear( );
while(m--)
{
scanf("%d%d%d",&s,&e,&tt);
add(s,e,tt);
add(e,s,tt);
}
while(t--)
{
scanf("%d%d%d",&s,&e,&tt);
add(s,e,-tt);
}
for( i=1;i<=n;i++)
{
if(dis[i]>=INF)
{
int ret=spfa(i);
if(ret==-1) break;
}
}
if(i<=n) printf("YES\n");
else printf("NO\n");
}
return 0;
}
