1122. Hamiltonian Cycle (25)汉密尔顿回路

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

哈密顿圈 所有点只经过一次的回路。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int mp[201][201];
int n,m,a,b,c,v[201],k;
int main()
{
    cin>>n>>m;
    for(int i = 0;i < m;i ++)
    {
        cin>>a>>b;
        mp[a][b] = mp[b][a] = 1;
    }
    cin>>m;
    for(int i = 0;i < m;i ++)
    {
        cin>>k;
        int flag = 1;
        if(k != n + 1)flag = 0;
        cin>>a;
        c = a;
        memset(v,0,sizeof(v));
        v[a] ++;
        for(int j = 1;j < k;j ++)
        {
            cin>>b;
            if(!mp[a][b])flag = 0;
            v[b] ++;
            if(v[b] == 2 && b != c || v[b] > 2)flag = 0;
            a = b;
        }
        if(b != c)flag = 0;
        if(flag)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}