题目地址

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 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

作者: CHEN, Yue
单位: 浙江大学
时间限制: 300 ms
内存限制: 64 MB

#include <cstdio>
#include <cstring>
#include<iostream>
#include <vector>
#include<math.h>
#include<string>
#include <algorithm>
#include<set>
using namespace std;
const int  maxn= 1010;  
bool G[maxn][maxn]={0};int circle[maxn];
int main(){
    int n,m,q1,q2,v1,v2;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>v1>>v2;
        G[v1][v2]=1;G[v2][v1]=1;
    }
    cin>>q1;
    for(int query=0;query<q1;query++){
        cin>>q2;
        int flag=1,repeat[maxn]={0};            //初始化
        if(q2<n+1) flag=0;              //长度小于n+1的；
        for(int i=0;i<q2;i++) cin>>circle[i];
        if(circle[0]!=circle[q2-1]) flag=0;             //开始结尾点不同的；
        for(int i=1;i<q2;i++){
            if(!G[circle[i]][circle[i-1]]){     //跟前一个点不连通的；
                flag=0;break;
            }
            repeat[circle[i]]++;                //点多次出现的；
            if(repeat[circle[i]]>1){
                flag=0;break;
            }
        }
        if(flag==1) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}