POJ 3259 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 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. 

Why POJ gg again ???

后来找了个别的OJ交的= =。。。

原题没有多组输入

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{    int u,v,w;} eage[6000];const int INF=0x3f3f3f3f;int dis[6000];int num;int n,m,w;void Add(int s,int e,int t){    num++;    eage[num].u=s;    eage[num].v=e;    eage[num].w=t;}int Bellman()//该算法用于图中有负权值的题目{    int i,j;    int flag;    for(i=0; i<=n; i++)    {        dis[i]=INF;    }    dis[1]=0;    flag=0;    for(i=1; i<=n; i++)    {        for(j=1; j<=num; j++)        {            if(dis[eage[j].v]>dis[eage[j].u]+eage[j].w)            {                dis[eage[j].v]=dis[eage[j].u]+eage[j].w;            }        }    }    for(i=1; i<=num; i++)    {        if(dis[eage[i].v]>dis[eage[i].u]+eage[i].w)            return 1;    }    return 0;}int main(){    int t;    while(~scanf("%d",&t))    {        while(t--)        {            scanf("%d%d%d",&n,&m,&w);            num=0;            while(m--)            {                int s,e,t;                scanf("%d%d%d",&s,&e,&t);                Add(s,e,t);                Add(e,s,t);            }            while(w--)            {                int s,e,t;                scanf("%d%d%d",&s,&e,&t);                Add(s,e,-t);            }            if(Bellman())printf("YES\n");            else printf("NO\n");        }    }    return 0;}