POJ 3259 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.

Source

USACO 2006 December Gold


/*
       大体的题意是这样的，一个农场主在农场里发现了很多虫洞，给予T个农场，每个农场有N块地，这N块地之间有M条路(双向边)，W个虫洞(单向边)，通过路会耗费一定的时间，而通过虫洞则会让时间倒流。题目问题是让求得农场主是否可以通过某一条路径从而看到还没有进虫洞的自己。即出洞时间大于出发时间，从而看到了自己。



       思路，用SPFA算法，邻接表存图。判断图中是否有负环出现，如果有，就说明农场主可以看到自己。

*/

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
const int INF = 0x3f3f3f3f;
using namespace std;

struct node
{
    int x;
    int sum;
    node *next;
}*head[1010];

int dis[1010];
int inq[1010];//标记是否存在于队列中
int cont[1010];//标记是否有重边

bool Spfa(int v0, int n)
{
    for(int i=0;i<=n;i++)
    {
        dis[i] = INF;inq[i] = 0;cont[i] = 0;
    }
    queue<int>Q;
    node *q;
    Q.push(v0);inq[v0]++;cont[v0]++;dis[v0]=0;
    while(!Q.empty())
    {
        int u = Q.front();Q.pop();inq[u]--;
        if(cont[u] > n)
            return true;
        q = head[u];
        while(q)
        {
            int v = q->x;
            if(dis[v] > dis[u] + q->sum)
            {
                dis[v] = dis[u] + q->sum;
                if(!inq[v])
                {
                    Q.push(v);inq[v]++;cont[v]++;
                }
            }
            q=q->next;
        }
    }
    return false;
}

int main()
{
    int t,n,m,w,s,e,b;
    node *p;
    scanf("%d",&b);
    while(b--)
    {
        memset(head,0,sizeof(head));
        scanf("%d %d %d",&n,&m,&w);
        for(int i=0;i<m;i++)//存储路
        {
            scanf("%d %d %d",&s,&e,&t);
            p = new node;
            p->sum = t;
            p->next = NULL;
            p->x = e;
            if(head[s] == NULL)
                head[s] = p;
            else
            {
                p->next = head[s];
                head[s] = p;
            }
            p = new node;
            p->sum = t;
            p->next = NULL;
            p->x = s;
            if(head[e] == NULL)
                head[e] = p;
            else
            {
                p->next = head[e];
                head[e] = p;
            }
        }
        for(int i=0;i<w;i++)//存储虫洞，单向边
        {
            scanf("%d %d %d",&s,&e,&t);
            p = new node;
            p->sum = -t;
            p->next = NULL;
            p->x = e;
            if(head[s] == NULL)
                head[s] = p;
            else
            {
                p->next = head[s];
                head[s] = p;
            }
        }
        if(Spfa(1,n))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}