PAT甲级 -- 1122 Hamiltonian Cycle (25 分)

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博客围绕哈密顿循环问题展开，要求判断给定的循环是否为哈密顿循环。介绍了输入输出规格，包含顶点数、边数等信息。还给出了示例输入输出，以及学习点，如判断节点是否多走、少走、成环，直接判断路径能否走通。

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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 ... Vn

where n is the number of vertices in the list, and Vi'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

22分：

又臭又长的代码：

#include <iostream>
#include <vector>
using namespace std;

int n, m, k; //顶点和边数
int visit[201] = {0};
int cnt = 0;
int g[201][201] = {0};

void dfs(int node, vector<int> a)
{
	visit[a[node]]++;
	for (int i = node+1; i < a.size(); i++)
	{
		if (g[a[node]][a[i]] == 1)
		{
			dfs(i,a);
			break;
		}
	}
}
int main()
{
	scanf("%d %d", &n, &m);
	for(int i = 0; i < m; i++)
	{
		int a,b;
		scanf("%d %d", &a, &b);
		g[a][b] = g[b][a] = 1;
	}

	scanf("%d", &k);
	for (int i = 0; i < k; i++)
	{
		int num;
		scanf("%d", &num);

		vector<int> ver(num,0); // 查询路径序列

		for (int j = 0; j < num; j++)
		{
			scanf("%d", &ver[j]);
		}
		
		if (ver[0] == ver[num-1] && num >= n) //回到起点
		{
			cnt = n-1;
			fill(visit, visit+201, 0);
			dfs(0, ver);
			for (int k = 1; k < ver.size()-1; k++)
			{
				if (visit[ver[k]] > 1 || visit[ver[k]] == 0)
				{
					cnt--;
				}
			}
			if (cnt != n-1)
			{
				printf("NO\n");
			}else
			{
				printf("YES\n");
			}
			
		}else
		{
			cnt = 0;
			printf("NO\n");
		}
		
	}
	return 0;
}

参考代码（柳神）：

学习点：

1. 判断节点是否多走、少走、或走成环
2. 判断这条路能不能走通，不需要dfs，直接判断即可..

#include <iostream>
#include <vector>
#include <set>
using namespace std;

int main()
{
	int n, m, num, k,g[201][201] = {0};
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; i++)
	{
		int a,b;
		scanf("%d%d", &a, &b);
		g[a][b] = g[b][a] = 1;
	}
	scanf("%d", &num);
	while(num--)
	{
		scanf("%d", &k);
		vector<int> v(k);
		set<int> s;
		for (int i = 0; i < k; i++)
		{
			scanf("%d", &v[i]);
			s.insert(v[i]);
		}
		int flag1 = 1, flag2 = 1;
		if (s.size() != n || v[0] != v[k-1] || k - 1 != n) flag1= 0;
		for (int i = 0; i < k - 1; i++)
		{
			if(g[v[i]][v[i+1]] == 0) flag2 = 0;
		}

		if (flag1 && flag2)
		{
			printf("YES\n");
		}else
		{
			printf("NO\n");
		}
	}

}