1122. Hamiltonian Cycle

最新推荐文章于 2022-11-19 15:31:31 发布

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最新推荐文章于 2022-11-19 15:31:31 发布

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本文探讨了图论中的汉密尔顿回路问题，即寻找一条简单回路，包含图中每一个顶点恰好一次。文章详细描述了一个判断给定路径是否构成汉密尔顿回路的算法实现，并提供了输入输出示例。

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<algorithm>
#include<vector>
using namespace std;
const int maxn=210;
bool vis[maxn];
int G[maxn][maxn]={0};
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		G[u][v]=G[v][u]=1;
	}
	int k;
	scanf("%d",&k);
	for(int i=0;i<k;i++)
	{
		int t,v,f=1,num=0;
		scanf("%d",&t);
		vector<int> path;
		memset(vis,0,sizeof(vis));
		for(int j=0;j<t;j++)
		{
			scanf("%d",&v);
			path.push_back(v);
			if(j<t-1&&vis[v]==false)
			{
				num++;
				vis[v]=true;
			}
			if(j>0)
			{
				int v1=path[j-1],v2=path[j];
				if(G[v1][v2]==0)f=0;
			}
		}
		//cout<<f<<endl;
		if(f==1)
		{
		if(t!=n+1||num!=n||path[0]!=path[t-1])
		{
			f=0;
		}
		/*else 
		{
		   for(int j=0;j<t-1;j++)
		   {
			   int l=j+2,r=t-2;
			   if(j==0)r--;
			   for(;l<=r;l++)
			   {
				   int v1=path[j],v2=path[l];
				   if(G[v1][v2]==1)
				   {
					   cout<<v1<<v2<<endl;
					   f=0;
					   break;
				   }
			   }
		   }
		}*/
		}
		if(f==1)printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}