POJ 3259 Wormholes spfa

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 toF (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 fromS 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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

思路  把虫洞的权值设为负值，判断是否存在负圈 

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int inf=0x3f3f3f3f;const int M=6000;const int N=510;struct sg{    int l,r;    int cost;    int next;}ev[M];int t;int n,m,w;int dis[N];int vis[N];int cmp;int head[M];int cnt[M];//前向星优化 http://www.layz.net/LAOJ/suanfa/s9-4.htmlvoid fuzhi(int a,int b,int c){    ev[cmp].l=a;    ev[cmp].r=b;//记录后继节点 相当于链表中的创建一个节点，并使得数据域先记录    ev[cmp].cost=c;    ev[cmp].next=head[a];//记录顶点节点的某一个边表节点的下标,    //相当于在链表中吧该边表节点的next指针先指向他的后继边表节点    head[a]=cmp++;}bool spfa(int st){    queue<int>q;    for(int i=0;i<=n;i++)    {        dis[i]=inf;    }    memset(vis,0,sizeof(vis));    memset(cnt,0,sizeof(cnt));    dis[st]=0;    q.push(st);    vis[st]=1;    cnt[st]++;    while(q.size())    {        int now=q.front();        q.pop();        vis[now]=0;        if(cnt[now]>=n)            return false;        for(int i=head[now];i!=-1;i=ev[i].next)        {            int nex=ev[i].r;            if(dis[nex]>dis[now]+ev[i].cost)            {                dis[nex]=dis[now]+ev[i].cost;                if(!vis[nex])                {                    vis[nex]=1;                    cnt[nex]++;                    if(!q.empty()&&dis[nex]<dis[q.front()])                    {                        q.push(nex);                    }                    else                    {                        q.push(nex);                    }                }            }        }    }    return true;}int main(){    scanf("%d",&t);    while(t--)    {        cmp=0;        int a,b,c;        memset(head,-1,sizeof(head));        scanf("%d%d%d",&n,&m,&w);        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            fuzhi(a,b,c);            fuzhi(b,a,c);        }        for(int i=0;i<w;i++)        {            scanf("%d%d%d",&a,&b,&c);            fuzhi(a,b,0-c);        }        if(spfa(1))            printf("NO\n");        else            printf("YES\n");    }    return 0;}