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 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

思路：

满足以下条件之一或几个就不是哈密顿环：

1. n不等于N+1；
2. 相邻两个结点不相连；
3. 头尾两个结点不一样；
4. 环没有覆盖所有结点。

C++：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int MAX = 210;
int n , m , q;
int mp[MAX][MAX]={0};
int visited[MAX];
vector<int> path;
int main(){
	cin>>n;
	cin>>m;
	for(int i=0;i<m;i++){
		int a , b;
		scanf("%d%d",&a,&b);
		mp[a][b]=mp[b][a]=1;
	}
	cin>>q;
	for(int j=0;j<q;j++){
		int flag=1;
		fill(visited,visited+MAX,0);
		path.clear();
		int num;
		scanf("%d",&num);
		for(int i=0;i<num;i++){
			int tempv;
			scanf("%d",&tempv);
			path.push_back(tempv);
		}
		if(*path.begin()!=*(path.end()-1)||num-1!=n) flag=0;
		else{
			for(int i=0;i<num-1;i++){
				if(visited[path[i]]==0&&mp[path[i]][path[i+1]]==1){
					visited[path[i]]=1;
				}
				else{
					flag=0;break;
				}
			}
		}
		if(flag==0){
			printf("NO\n");
		}
		else{
			printf("YES\n");
		}
	}
	return 0;
}