poj 3259 Wormholes

Wormholes

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 37102

Accepted: 13621

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 Tseconds.

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.

题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前;

解决方法：用spfa写即可，虫洞单向，道路双向！，并且单向为负路径！！

代码：

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
#include<queue>
#define INF 0x3f3f
#include<string.h>
#define MAXN 2000
#define MAXM 20010
struct Edge
{
 int from,to,val,next;
}edge[MAXM];
int dist[MAXN],vis[MAXN],head[MAXN],used[MAXN];
int n,m,top=0,k,flag;
void init()//初始化 。，注意初始化尽量在函数里面，我这个写的不太好，但这个题貌似不影响！ 
{
    memset(vis,0,sizeof(vis));
    memset(dist,INF,sizeof(dist));
    memset(head,-1,sizeof(head));
    memset(used,0,sizeof(used));
}
void addEdge(int u,int v,int w)//建邻接表 
{
    edge[top].from=u;
    edge[top].to=v;
    edge[top].val=w;
    edge[top].next=head[u];
    head[u]=top++;
}
int spfa(int sx)
{
    flag=0;
    queue<int>Q;
    Q.push(sx);
    dist[sx]=0;
    vis[sx]=1;//标记 
    used[sx]=1;
    while(!Q.empty() )
    {
        int u=Q.front();
        Q.pop() ;
        vis[u]=0;//去标记 
        for(int i=head[u];i!=-1;i=edge[i].next )
        {
            int v=edge[i].to ;
            if(dist[v]>dist[u]+edge[i].val )
            {
                dist[v]=dist[u]+edge[i].val ;
                if(!vis[v])
                {
                    vis[v]=1;//标记 
                    Q.push(v); 
                    used[v]++;
                    if(used[v]>n)//看是否大于节点数 ，大于则出现负环 
                    {
                        flag=1;
                        return 1;
                    }
                }
            }
        }
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        top=0;
        int u,v,w;
    scanf("%d%d%d",&n,&m,&k);
    while(m--)
    {
        scanf("%d%d%d",&u,&v,&w);//取路，双向 
        addEdge(u,v,w);
        addEdge(v,u,w);
    }   
    while(k--)//单向，并且为负 
    {
        scanf("%d%d%d",&u,&v,&w);
        addEdge(u,v,-w);
    }
      int s=spfa(1);
      if(s)
      printf("YES\n");
      else
      printf("NO\n");
    }
    return 0;
}