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poj3259

发布时间:2022-07-27 13:55:50
                                                                                       Wormholes
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 24910   Accepted: 8883
Descriptionpoj3259

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..NM (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, FF farm descriptions follow. 
Line 1 of each farm: Three space-separated integers respectively: NM, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (SET) 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 (SET) 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

Hint

For farm 1, FJ cannot travel back in time. 
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<cstring>
#include<cstdio>
#include<string>
using namespace std;
const int inf = 10000000;
const int maxn =3000;
int dist[maxn],i,j,n,m,w,u,v,e,x,y,z,cnt,k;
struct node
{
    int u,v,w;
}edge[maxn*2];
int bellman()
{
    for(i=0;i<n;i++)
    {
        dist[i] = inf;
    }
    dist[x] = 0;
    for(k=1;k<n;k++)
    {
        for(i=0;i<cnt;i++)
        {
            if(dist[edge[i].u]!=inf && edge[i].w+dist[edge[i].u]<dist[edge[i].v])
            {
                dist[edge[i].v] = edge[i].w + dist[edge[i].u];
            }
        }
    }
    for(i=0;i<cnt;i++)
    {
            if(edge[i].w+dist[edge[i].u]<dist[edge[i].v])
            {
                return 1;
            }
            
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&m,&w);
        cnt =0;
        for(i=0;i<m;i++)
        {
            scanf("%d %d %d",&u,&v,&e);
            edge[cnt].u= u;
            edge[cnt].v = v;
            edge[cnt++].w = e;
            edge[cnt].u = v;
            edge[cnt].v = u;
            edge[cnt++].w=e;
        }
        for(i=0;i<w;i++)
        {
            scanf("%d %d %d",&x,&y,&z);
            edge[cnt].u = x;
            edge[cnt].v = y;
            edge[cnt++].w = -z;
        }
        int num = bellman();
        if(num)
        {
            printf("YES
");
        }
        else
            printf("NO
");
    }
    return 0;
}

/template/Home/DawnNew/PC/Static

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