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.
t译文：农夫约翰在探索他的许多农场，发现了一些惊人的虫洞。虫洞是很奇特的，因为它是一个单向通道，可让你进入虫洞的前达到目的地！他的N（1≤N≤500）个农场被编号为1..N，之间有M（1≤M≤2500）条路径，W（1≤W≤200）个虫洞。作为一个狂热的时间旅行FJ的爱好者，他要做到以下几点：开始在一个区域，通过一些路径和虫洞旅行，他要回到最开时出发的那个区域出发前的时间。也许他就能遇到自己了:)。为了帮助FJ找出这是否是可以或不可以，他会为你提供F个农场的完整的映射到（1≤F≤5）。所有的路径所花时间都不大于10000秒，所有的虫洞都不大于万秒的时间回溯。
输入
第1行：一个整数F表示接下来会有F个农场说明。
每个农场第一行：分别是三个空格隔开的整数：N，M和W
第2行到M+1行：三个空格分开的数字（S，E，T）描述，分别为：需要T秒走过S和E之间的双向路径。两个区域可能由一个以上的路径来连接。
第M +2到M+ W+1行：三个空格分开的数字（S，E，T）描述虫洞，描述单向路径，S到E且回溯T秒。
输出
F行，每行代表一个农场
每个农场单独的一行，” YES”表示能满足要求，”NO”表示不能满足要求。
NO
YES
思路：
虫洞可以让 你回到过去，就相当于一个负权。然后判断是否存在负环，即走走走。。。然后又走回去了
1 #include <stdio.h> 2 #include <algorithm> 3 #include <iostream> 4 #include <stdbool.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <string.h> 8 #include <math.h> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <map> 13 14 #define INF 0x3f3f3f3f 15 #define LL long long 16 #define MAXN 1000005 17 using namespace std; 18 19 typedef struct Edge{ 20 int u,v; 21 int cost; 22 }Edge; 23 24 int nodenum,edgenum; 25 int hole; 26 int T; 27 Edge edge[MAXN]; 28 int dist[MAXN],pre[MAXN]; 29 30 bool Bellman_Ford() 31 { 32 for (int i=1;i<=nodenum;i++) 33 dist[i] = INF; 34 dist[1] = 0; 35 for (int i=1;i<nodenum;i++) 36 { 37 for (int j=1;j<=edgenum;j++) 38 { 39 if (dist[edge[j].v] > dist[edge[j].u]+edge[j].cost) 40 { 41 dist[edge[j].v] = dist[edge[j].u]+edge[j].cost; 42 pre[edge[j].v] = edge[j].u; 43 } 44 } 45 } 46 bool flag = true; 47 for (int i=1;i<=edgenum;i++) 48 { 49 if (dist[edge[i].v]>dist[edge[i].u]+edge[i].cost) 50 { 51 flag = false; 52 break; 53 } 54 } 55 return flag; 56 } 57 58 void print_path(int root) 59 { 60 while (root!=pre[root]) 61 { 62 printf("%d->",root); 63 root = pre[root]; 64 } 65 if (root == pre[root]) 66 printf("%d ",root); 67 } 68 69 70 int main() 71 { 72 //freopen("../in.txt","r",stdin); 73 scanf("%d",&T); 74 while (T--) 75 { 76 scanf("%d%d%d",&nodenum,&edgenum,&hole); 77 memset(edge,0,sizeof(edge)); 78 int i=0; 79 int a,b,c; 80 for (int j=1;j<=edgenum;j++) 81 { 82 scanf("%d%d%d",&a,&b,&c); 83 i++; 84 edge[i].u = a; 85 edge[i].v = b; 86 edge[i].cost = c; 87 i++; 88 edge[i].u = b; 89 edge[i].v = a; 90 edge[i].cost = c; 91 } 92 for (int j=1;j<=hole;j++) 93 { 94 scanf("%d%d%d",&a,&b,&c); 95 i++; 96 edge[i].u = a; 97 edge[i].v = b; 98 edge[i].cost = -c; 99 } 100 edgenum = i; 101 if (Bellman_Ford()) 102 printf("NO "); 103 else 104 printf("YES "); 105 } 106 return 0; 107 }