Wormholes

原创于 2018-08-14 21:36:40 发布 · 253 阅读

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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 requiresT 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

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.

题意：每组测试数据第一行表示农场里N块地，M条路连接两块地，W个虫洞，然后接下来M行的每行是三个数S，E，T代表:S和E之间的双向路径，需要T秒才能穿过，接下来W行，每行三个数S，E，Ts，表示从S到E的·单向路径，可让旅行者返回Ts秒虫洞使条单向路，会在你进入虫洞之前把你传送到目的地，就是当你过去的时候时间会倒退Ts。求会不会在最初离开之前的某个时间之前回到起点。

思路：这个题就是问我们是否有负权回路。在输遇到虫洞的三个数时把时间输成负数就行了，然后用SPFA算法。代码如下：


#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define maxx 100000
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
    int to;
    int next;
    int w;
} e[maxx];
int flag;
int dis[maxx];
int head[maxx],cn[maxx],sum[maxx],vis[maxx];
int cnt,m,n;
void add(int x,int y,int d)
{
    e[cnt].to=y;
    e[cnt].w=d;
    e[cnt].next=head[x];
    head[x]=cnt++;
}
void spfa()
{
    queue<int>q;
    q.push(1);
    dis[1]=0;
    vis[1]=1;
    while(!q.empty())
    {
        int u=q.front();
        vis[u]=0;
        q.pop();
        for(int i=head[u]; i!=-1; i=e[i].next)
        {
            int v=e[i].to;
            if(dis[v]>dis[u]+e[i].w)
            {
                dis[v]=dis[u]+e[i].w;
                if(!vis[v])
                {
                    sum[v]++;
                    vis[v]=1;
                    q.push(v);
                    if(sum[v]>=n)
                    {
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(flag)
            break;
    }
}
int main()
{
    int T,c;
   scanf("%d",&T);
    while(T--)
    {
        memset(dis,inf,sizeof(dis));
        memset(head,-1,sizeof(head));
        memset(cn,0,sizeof(cn));
        memset(sum,0,sizeof(sum));
        memset(vis,0,sizeof(vis));
        cnt=0;
        flag=0;
        scanf("%d%d%d",&n,&m,&c);
        int x,y,z;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
            add(y,x,z);
        }
        for(int i=0; i<c; i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,-z);
        }
        spfa();
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}