F - Wormholes（判断是否存在负环）

##题目描述
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

题目大意：第一行输入一个数是表示几组测试数据
第二行三个数 N(点的个数),M(正边的个数),W(负边的个数) 注意：正边为双向的，负边为单向的。
然后 M行u,v,w;
再然后W行u,v,w;
求这个图是不是存在负环。有 YES 没NO。

题解：思路与判断是否存在正环是一样的，我们用结构体来存u， v， w，松弛一遍判断就行

#include<iostream>
#include<stdio.h>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;
struct p{int x, y, z;
}arr[1000005];
int dis[1000005];
int n, k;
bool djs()
{memset(dis, INF, sizeof dis);  //dis数组在判断负圈是否存在时，不用初始化，dis[1] = 0;                   //当然，初始化了也没有什么影响for(int i = 1; i < n; i++){  //n-1条边for(int j = 0; j < k; j++){  //松弛一边dis[arr[j].y] = min(dis[arr[j].y], dis[arr[j].x] + arr[j].z);}}for(int j = 0; j < k; j++){//如果还存在这种情况就肯定存在负圈if(dis[arr[j].y] > dis[arr[j].x] + arr[j].z)return true;}return false;
}
int main()
{int t;cin >> t;int a, b, c;int m, w;while(t--){k = 0;scanf("%d%d%d", &n, &m, &w);for(int i = 1; i <= m; i++){scanf("%d%d%d", &a, &b, &c);arr[k].x = a;arr[k].y = b;arr[k++].z = c;arr[k].x = b;arr[k].y = a;arr[k++].z = c;}for(int i = 1; i <= w; i++){scanf("%d%d%d", &a, &b, &c);arr[k].x = a;arr[k].y = b;arr[k++].z = -c;}if(djs())cout << "YES" << endl;else cout << "NO" << endl;}return 0;
}