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

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.

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
         
          #include
          
           #include
           
            using namespace std;#define N 550#define INF 0xffffffstruct node{    int y, w;}P[N];vector
            
              G[N];int v[N];int spfa(int s){    queue
             
               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。