The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".

In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<N≤200), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format Vertex1 Vertex2, where the vertices are numbered from 1 to N. The next line gives a positive integer K which is the number of queries, followed by K lines of queries, each in the format:

n V​1​​ V​2​​ ... V​n​​

where n is the number of vertices in the list, and V​i​​'s are the vertices on a path.

Output Specification:

For each query, print in a line YES if the path does form a Hamiltonian cycle, or NO if not.

Sample Input:

6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1

Sample Output:

YES
NO
NO
NO
YES
NO
  
 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <map>
 5 #include <vector>
 6 #include <set>
 7 using namespace std;
 8 int n,m,k;
 9 int dis[201][201];
10 int path[1000],vis[201];
11 int main(){
12     scanf("%d %d",&n,&m);
13     for(int i=0;i<m;i++){
14         int c1,c2;
15         scanf("%d %d",&c1,&c2);
16         dis[c1][c2]=1;
17         dis[c2][c1]=1;
18     }
19     scanf("%d",&k);
20     int min=99999999,mini=0;
21     for(int i=1;i<=k;i++){
22         int flag=0;
23         int total=0;
24         int nn;
25         fill(vis,vis+201,0);
26         scanf("%d",&nn);
27         for(int j=0;j<nn;j++){
28              scanf("%d",&path[j]);
29              vis[path[j]]++;
30         }
31         for(int j=1;j<=n;j++){
32              if(vis[j]==0) flag=1;
33         }
34         if(nn!=n+1 || path[0]!=path[nn-1]) flag=1;
35         for(int j=1;j<nn;j++){
36               if(dis[path[j]][path[j-1]]==0){
37                   flag=1;
38                   break;
39             }
40           }
41           if(flag==1) printf("NO\n");
42           else{
43               printf("YES\n");
44         }
45     }
46 }
View Code

注意点：挺简单的一道题，就按题目意思实现一下，判断给的路径是否是一个环，这个环有没有包含所有节点，并且是联通的，点除了起点都只经过一次。和1150 Travelling Salesman Problem （25 分）差不多，比他更简单一些