1122. Hamiltonian Cycle (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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<stdio.h>
#include<algorithm>
using namespace std;
int N,m;
int adj[210][210];
int judge(){
	int n,i;
	scanf("%d",&n);	
	int a[n];
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	if(n!=N+1){//满足个数为N+1 
		return 0;
	}
	if(a[0]!=a[n-1]){//首尾同一个 
		return 0;
	}
	for(i=0;i<n-1;i++){ 
		int now=a[i],next=a[i+1];
		if(adj[now][next]==0){//相邻两点是否有边 
			return 0;
		}
	}
	sort(a,a+n-1);
	for(i=0;i<n-2;i++){//测试点2 
		if(a[i]==a[i+1]){//每个点出现1次 
			return 0;
		}
	}
	return 1;
}
int main(){
	int c1,c2,i,k;
	scanf("%d %d",&N,&m);
	for(i=0;i<m;i++){
		scanf("%d %d",&c1,&c2);
		adj[c1][c2]=adj[c2][c1]=1;
	}
	scanf("%d",&k);
	for(i=0;i<k;i++){
		if(judge()){
			printf("YES\n");
		}
		else{
			printf("NO\n");
		}
	}
}