PAT (Advanced Level) Practise 1122 Hamiltonian Cycle (25)（哈密顿圈）

最新推荐文章于 2020-03-24 11:39:53 发布

原创最新推荐文章于 2020-03-24 11:39:53 发布 · 257 阅读

0 ·

CC 4.0 BY-SA版权

----水题同时被 3 个专栏收录

234 篇文章

订阅专栏

----其他

216 篇文章

订阅专栏

PAT/PTA甲级

174 篇文章

订阅专栏

本文介绍了一道关于哈密顿圈的问题，通过输入点、边和路径，判断给定路径是否构成哈密顿圈，并提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题意：有n个点，m条边，q次询问，给出一个走的序列，问是不是哈密顿圈，哈密顿圈满足：序列首尾相同，序列长度 = 顶点个数 + 1，除首尾外，不会再有相同的顶点，序列中相邻的两个顶点要求有边相连。

解题思路：水题，所有条件判一下就好，不过题目数据好像有点怪，我数组205的有一个点过不了，2050就过了

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <cmath>  
#include <map>  
#include <cmath>  
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long  
const int INF = 0x3f3f3f3f;

int n, m, q, k, a[2050];
int edge[2050][2050], visit[2050];

int main()
{
	scanf("%d%d", &n, &m);
	int u, v;
	memset(edge, 0, sizeof edge);
	for (int i = 1; i <= m; i++)
	{
		scanf("%d%d", &u, &v);
		edge[u][v] = edge[v][u] = 1;
	}
	scanf("%d", &q);
	while (q--)
	{
		memset(visit, 0, sizeof visit);
		scanf("%d", &k);
		int cnt = 0, flag = 1;
		for (int i = 1; i <= k; i++)
		{
			scanf("%d", &a[i]);
			if (!visit[a[i]]) visit[a[i]] = 1, cnt++;
		}
		if (a[1] != a[k] || cnt != n || k != (n + 1)) flag = 0;
		for (int i = 2; i <= k; i++)
			if (!edge[a[i - 1]][a[i]]) flag = 0;
		if (flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}