Wormholes

Wormholes Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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 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.

意思就是看看从一个点出发经过多个边权值（肯定有负的，回到自身的时候是不是负的，那样就是时光倒流了，本题是看1那点，能不能时光倒流

题目大意：

农民约翰在农场散步的时候发现农场有大量的虫洞，这些虫洞是非常特别的因为它们都是单向通道，为了方便现在把约翰的农田划分成N快区域，M条道路，W的虫洞。

约翰是时空旅行的粉丝，他希望这样做，在一个区域开始，经过一些道路和虫洞然后回到他原来所在的位置，这样也许他就能见到他自己了。

穿越虫洞会回到以前。。。。。（穿越者约翰）

/////////////////////////////////////////////////////////////

很明显这是一个很扯淡的故事，不过为了出一道带负权值的题目也是难为了出题人，迪杰斯特拉算法不能处理带有负权值的问题，佛洛依德倒是可以，不过复杂度很明显太高，所以spfa是不错的选择，只需要判断出发点时间是否变小.

#include<iostream>
#include<vector>
#include<queue>

using namespace std;

#define N 550
#define INF 0xffffff

struct node
{
    int y, w;
}P[N];

vector<node> G[N];
int v[N];

int spfa(int s)
{
    queue<int> Q;

    Q.push(s);

    while(Q.size())
    {
        s = Q.front();
        Q.pop();

        int len = G[s].size();

        for(int i = 0; i < len; i++)
        {
            node q = G[s][i];

            if(v[s] + q.w < v[q.y])
            {
                v[q.y] = v[s] + q.w;
                Q.push(q.y);
            }


        }
        if(v[1] < 0)
            return 1;
    }

    return 0;
}
int main()
{
    int t, n, m, w, s, e, f;

     cin >> f;

     while(f--)
     {
         for(int i = 1; i < N; i++)
         {
             v[i] = INF;
             G[i].clear();
         }

         v[1] = 0;

         cin >> n >> m >> w;
         node p;

         for(int i = 0; i < m; i++)
         {
             cin >> s >> e >> t;
             p.y = e, p.w = t;
             G[s].push_back(p);
             p.y = s;
             G[e].push_back(p);
         }
         for(int i = 0; i < w; i++)
         {
             cin >> s >> e >> t;
             p.y = e, p.w = -t;
             G[s].push_back(p);
         }
         int ans = spfa(1);

         if(ans)
            cout << "YES" << endl;
         else
            cout << "NO" << endl;
     }
     return 0;
}

SPFA 在形式上和宽度优先搜索非常类似，不同的是宽度优先搜索中一个点出了队列就不可能重新进入队列，但是SPFA中一个点可能在出队列之后再次被放入队列，也就是一个点改进过其它的点之后，过了一段时间可能本身被改进，于是再次用来改进其它的点，这样反复迭代下去。