Wormholes

                     Wormholes 

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
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：约翰的牧场有N个区域，这些区域直接有一些道路连接，另外约翰还发现一些虫洞，可以让他从S到E时间倒退T秒，问他能否走过若干路径之后回到原点看到出发前的自己？

思路：把带有虫洞的边看做负权边，若能看到出发前的自己即能找到负权环，用spfa判断一个节点重复n次以上出现在队列中即出现负权环。

code：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint map[505][505];int first[1000010],dis[1000010];int next[1000010],point[1000010],w[1000010],q[1000010];int cnt[1000010];bool f[1000010];int n,m,ww,tot;void add(int x,int y,int c){    tot++;    w[tot]=c;    next[tot]=first[x];    first[x]=tot;    point[tot]=y;}bool spfa(int s){    memset(dis,INF,sizeof(dis));    memset(f,0,sizeof(f));    memset(cnt,0,sizeof(cnt));    int l,r,x,y,k;    l=100000;    r=l;    dis[s]=0;    q[l]=s;    f[s]=true;    while (l<=r)     {         x=q[l];         if(cnt[x] >= n)            return false;         l++;         k=first[x];         f[x]=false;         while(k!=0)           {              y=point[k];              if(dis[y]>dis[x]+w[k])                {                    dis[y]=dis[x]+w[k];                    if (!f[y])                      {                          cnt[y]++;                          if (dis[y]<dis[q[l]])                            {                                l--;                                q[l]=y;                                f[y]=true;                            } else                            {                                r++;                                q[r]=y;                                f[y]=true;                            }                      }                }              k=next[k];           }     }    return true;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        tot = 0;        memset(first,0,sizeof(first));        scanf("%d%d%d",&n,&m,&ww);        int x,y,z;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&x,&y,&z);            add(x,y,z);            add(y,x,z);        }        for(int i=1;i<=ww;i++)        {            scanf("%d%d%d",&x,&y,&z);            add(x,y,-z);            add(x,y,-z);        }        if(spfa(1))           printf("NO\n");        else           printf("YES\n");    }    return 0;}