1122. Hamiltonian Cycle (25)

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 V₁ V₂ ... 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

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<algorithm>  
#include<queue>  
#include<stack>  
#include<set>
using namespace std;
int main(){
	int n,m;
	int map[300][300]={0};
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		map[a][b]=map[b][a]=1;
	}
	int k;
	scanf("%d",&k);
	while(k--){
		int len;
		int a[1000];
		scanf("%d",&len);
		int v[1000]={0};
		for(int i=0;i<len;i++){
			scanf("%d",&a[i]);
			v[a[i]]++;
		}
		if(a[0]!=a[len-1]||len!=n+1){
			printf("NO\n");
			continue;
		}
		int i;
		for(i=1;i<=n;i++){
		   		if(v[i]!=1&&(i!=a[0]||i!=a[len-1])){
					break;
	    	}
	    }
	    if(i!=n+1) printf("NO\n");
		else{
			int pro=a[0];
			int flag=0;
			for(int i=1;i<len;i++){
				if(map[a[i]][pro]){
					pro=a[i];
				}
				else{
					flag=1;
					break;
				}
			}
			if(flag) printf("NO\n");
			else printf("YES\n");
		} 
	}
	
    return 0;  
}