G - 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 backT 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.

题目意思; 有一个农场中有农田和虫洞,虫洞可以倒退一段时间,判断一个人能否回到出发前;

解题思路:判断是否存在负权回路;

 1 #include<iostream>
 2 #include <string.h>
 3 #include <math.h>
 4 #include <stdio.h>
 5 using namespace std;
 6
 7 const int MAX = 10000 + 100;
 8 long long MAX1 = 1e10;
 9 long long dist[MAX];
10 int add;
11 int n1,n2,n3;
12
13 struct S
14 {
15     int a,b;
16      int len;
17 };
18 S Map[MAX];
19 bool B()
20 {
21     for(int i = 1;i <= n1;i++)
22         dist[i] = MAX1;
23     dist[1] = 0;
24
25     for(int i =1;i <=n1;i++)
26     {
27         bool flag = false;
28         for(int i = 0;i < add;i++)
29         {
30             if( dist[ Map[i].b ] > dist[Map[i].a] + Map[i].len )
31             {
32                 dist[ Map[i].b ] = dist[Map[i].a] + Map[i].len;
33                 flag = true;
34             }
35
36         }
37         if(!flag)
38             break;
39     }
40
41     for(int i = 0;i <add;i++)
42         if(dist[ Map[i].b ] > dist[Map[i].a] + Map[i].len)
43             return true;
44     return false;
45
46 }
47
48
49 int main()
50 {
51     int N;
52     cin>>N;
53
54     while(N--)
55     {
56         cin>>n1>>n2>>n3;
57
58         add = 0;
59         for(int i =1;i <= n2;i++)
60         {
61             int x,y,len;
62             cin>>x>>y>>len;
63             Map[add].a = x;Map[add].b=y;Map[add++].len = len;
64             Map[add].a = y;Map[add].b=x;Map[add++].len = len;
65
66         }
67         for(int i = 1;i <= n3;i++)
68         {
69             int x,y,len;
70             cin>>x>>y>>len;
71              Map[add].a = x;Map[add].b=y;Map[add++].len = -len;
72         }
73
74         if( B() )
75             cout<<"YES"<<endl;
76         else
77             cout<<"NO"<<endl;
78
79     }
80
81
82     return 0;
83 }