PAT甲级 1122. Hamiltonian Cycle (25)

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

————————————————————————————————

题目的意思是给出n个点m条边，对于每次查询判断是否是哈密顿路径

思路：根据查询，分别判断点数，首尾是否一致，是否每个点都遍历，路径是否联通

若均符合则YES否则NO

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

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

int mp[500][500];
int a[500];
int main()
{
    int n,m,x,y,q,k;
    scanf("%d%d",&n,&m);
    memset(mp,0,sizeof mp);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&x,&y);
        mp[x][y]=mp[y][x]=1;
    }
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d",&k);
        for(int i=0;i<k;i++)
            scanf("%d",&a[i]);
            if(n==1&&k==1)
            {
                printf("YES\n");
                continue;
            }
        if(k!=n+1||a[0]!=a[k-1])
        {
            printf("NO\n");
            continue;
        }
        int flag=0;
      set<int>s;
      s.insert(a[0]);
        for(int i=1;i<k;i++)
        {
            if(mp[a[i]][a[i-1]]==0)
        {
            flag=1;
            break;
        }
        s.insert(a[i]);
        }
            if(s.size()!=n)
                flag=1;
        printf("%s\n",flag==0?"YES":"NO");

    }
    return 0;
}