1122 Hamiltonian Cycle （25 分）

1122 Hamiltonian Cycle （25 分）

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 Kwhich 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

题意：判断是否为Hamiltonian Cycle：条件是简单回路（除最后一个结点以外只出现一次），并且相邻两点之间连通。

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-28-00.13.48
 6 * Description : A1122
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=210;
20 const int INF=1000000000;
21 int G[maxn][maxn];
22 int n,m,k;
23 int main(){
24 #ifdef ONLINE_JUDGE
25 #else
26     freopen("1.txt", "r", stdin);
27 #endif
28     scanf("%d%d",&n,&m);
29     fill(G[0],G[0]+maxn*maxn,INF);
30     int u,v;
31     for(int i=0;i<m;i++){
32         scanf("%d%d",&u,&v);
33         G[u][v]=G[v][u]=1;
34     }
35     scanf("%d",&k);
36     int t,a;
37     for(int i=0;i<k;i++){
38         scanf("%d",&t);
39         vector<int> temp;
40         bool vis[maxn]={false};
41         bool flag=true;
42         for(int j=0;j<t;j++){
43             scanf("%d",&a);
44             temp.push_back(a);
45         }
46         if(temp[0]!=temp[temp.size()-1]){
47             printf("NO\n");
48             continue;
49         }
50         for(int x=0;x<temp.size()-1;x++){
51             if(G[temp[x]][temp[x+1]]!=1 || vis[temp[x]]==true){
52                 printf("NO\n");
53                 flag=false;
54                 break;
55             }
56             else{
57                 vis[temp[x]]=true;
58             }
59         }
60         if(flag==false) continue;
61         for(int j=1;j<n;j++){
62             if(vis[j]==false){
63                 printf("NO\n");
64                 flag=false;
65                 break;
66             }
67         }
68         if(flag) printf("YES\n");
69     }
70     return 0;
71 }