#1122. Hamiltonian Cycle【图论 + 模拟】

原题链接

Problem Description：

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 $N$ ($2<N\le 200$), 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:

$n{V}_1{V}_2\dots V_n$

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

Problem Analysis:

前置知识：

• 哈密顿路径：由指定的起点前往指定的终点，途中经过所有其他节点且只经过一次的路径

• 哈密顿回路：闭合的哈密顿路径称作哈密顿回路（Hamiltonian cycle）

• 哈密顿图：包含哈密顿回路的图

本题只需要从定义出发判断给定的序列是否是哈密顿回路，即只需满足以下四个条件，即使哈密顿回路的路径序列：

1. 序列总点数必须等于 $n+1$
2. 首尾必须是同一个点
3. 每个点能且只能出现一次，起始点除外，可以出现两次
4. 每个点的前驱必须是合法的，即不能出现不存在的边

Code

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 210;

int n, m;
bool g[N][N], st[N];
int nodes[N * 2];

bool check(int cnt)
{
    
    
    if (nodes[0] != nodes[cnt - 1] || cnt != n + 1) return false;
    
    memset(st, 0, sizeof st);
    for (int i = 0; i < cnt; i ++ )
    {
    
    
        st[nodes[i]] = true;
        if (!g[nodes[i]][nodes[i + 1]])
            return false;
    }
    
    for (int i = 1; i <= n; i ++ )
        if (!st[i])
            return false;
    return true;
}

int main()
{
    
    
    cin >> n >> m;
    while (m -- )
    {
    
    
        int a, b;
        cin >> a >> b;
        g[a][b] = g[b][a] = true;
    }
    
    int k;
    cin >> k;
    while (k -- )
    {
    
    
        int cnt;
        cin >> cnt;
        for (int i = 0; i < cnt; i ++ ) cin >> nodes[i];
        
        if (check(cnt)) puts("YES");
        else puts("NO");
    }
    return 0;
}