【PAT】1122. Hamiltonian Cycle (25)

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

分析：按哈密顿回路的定义走一圈回路即可

#include 
#include 
using namespace std;
int g[201][201];
int visited[201];
int main()
{
	freopen("test.txt","r",stdin);
	int n,m,k;
	cin>>n>>m;
	while(m--){
		int x,y;
		cin>>x>>y;
		g[x][y]=g[y][x]=1;
	}
	cin>>k;
	while (k--){
		fill(visited,visited+201,0);
		int s,t;
		cin>>s;
		if (s!=n+1) {
			while (s--) cin>>t;
			cout<<"NO\n";
		}else{
			cin>>t;
			visited[t]+=1;
			--s;
			int x,y;
			x=t;
			while (s--){
				cin>>y;
				if (g[x][y]==1) visited[y]+=1;
				x=y;
			}
			if (count(visited,visited+201,1)==n-1 && y==t) cout<<"YES\n";
			else cout<<"NO\n";
		}		
	}
	return 0;
}