一种路是双向的，路的长度是正值；另一种路是单向的，路的长度是负值； 如果有负环输出YES；否则输出NO；不同的路可能有相同的起点和终点：必须用邻接表
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.
Sponsor
1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 #include<algorithm>
6 #include<cmath>
7 #include<queue>
8 #include<vector>
9 #include<map>
10 #include<string>
11 #define LL long long
12 #define eps 1e-8
13 using namespace std;
14 const int inf = 0x3f3f3f3f;
15 int n,m,w,a,b,c,t,tot,ans;
16 struct node{
17 int l,r,num,nex;
18 }e[10010];
19 int head[550],vis[550],dis[550],sum[550];
20
21 void init()
22 {
23
24 memset(head,-1,sizeof(head));
25 tot=0;
26 ans=0;
27 }
28
29 void add(int l,int r,int num)
30 {
31 e[tot].l=l,e[tot].r=r,e[tot].num=num;
32 e[tot].nex=head[l];
33 head[l]=tot++;
34 }
35
36 int spfa(int x)
37 {
38 queue<int >q;
39 while(!q.empty())
40 q.pop();
41 memset(dis,inf,sizeof(dis));
42 memset(vis,0,sizeof(vis));
43 memset(sum,0,sizeof(sum));
44 q.push(x);
45 vis[x]=1;
46 dis[x]=0;
47 sum[x]=1;
48 while(!q.empty())
49 {
50 int now=q.front(); q.pop();
51 vis[now]=0;
52 for(int i=head[now];i!=-1;i=e[i].nex)
53 {
54 int to=e[i].r;
55 if(dis[to]>dis[now]+e[i].num)
56 {
57 dis[to]=dis[now]+e[i].num;
58 if(!vis[to])
59 {
60 sum[to]++;
61 if(sum[to]>n) return 1;
62 q.push(to);
63 vis[to]=1;
64 }
65 }
66 }
67 }
68 return 0;
69 }
70 int main()
71 {
72 scanf("%d",&t);
73 while(t--)
74 {
75 scanf("%d%d%d",&n,&m,&w);
76 init();
77 while(m--)
78 {
79 scanf("%d%d%d",&a,&b,&c);
80 add(a,b,c);
81 add(b,a,c);
82 }
83 while(w--)
84 {
85 scanf("%d%d%d",&a,&b,&c);
86 add(a,b,-c);
87 }
88 for(int i=1;i<=n;i++)
89 {
90 if(spfa(i)){
91 ans=1;
92 break;//否则超时
93 }
94 }
95 if(ans) printf("YES\n");
96 else printf("NO\n");
97 }
98 }
