PAT甲级——A1122 Hamiltonian Cycle【25】

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), 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​​ ... 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

第一，遍历点K一定等于N+1，因为既要遍历且一次所有顶点，而且回到起点
第二，遍历的最后一个点一定是起点
使用visit记录顶点是否遍历过了，graph记录路径是否能行

#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int n, m, k, a, b, start, graph[205][205];
    bool visit[205];//记录每个顶点遍历一次
    fill(graph[0], graph[0] + 25 * 205, -1);
    cin >> n >> m;
    while (m--)
    {
        cin >> a >> b;
        graph[a][b] = graph[b][a] = 1;
    }
    cin >> m;
    while (m--)
    {
        cin >> k;
        bool flag = true;
        fill(visit, visit + 205, false);
        for (int i = 0; i < k; ++i)
        {
            cin >> b;
            if (flag == false || k != n + 1)//遍历所有的顶点并回到起点，则一定走过n+1个点
            {
                flag = false;
                continue;
            }
            if (i == 0)
                start = b;//记录起点
            else if (graph[a][b] != 1)//此路不通
                flag = false;
            else if (i == k - 1 && b != start)//最后一个点不是起点
                flag = false;
            else if (i != k - 1 && visit[b] != false)//除了最后一次重复遍历起点，出现了其他点重复遍历，
                flag = false;
            else
                visit[b] = true;//遍历过
            a = b;//记录前一个点
        }
        if (flag)
        {
            for (int i = 1; i <= n && flag == true; ++i)
                if (visit[i] == false)//存在没有遍历的顶点
                    flag = false;
            if (flag)
                cout << "YES" << endl;
        }
        if (flag == false)
            cout << "NO" << endl;
    }
    return 0;    
}