1122. 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< 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:

6 106 23 41 52 53 14 11 66 31 24 567 5 1 4 3 6 2 56 5 1 4 3 6 29 6 2 1 6 3 4 5 2 64 1 2 5 17 6 1 3 4 5 2 67 6 1 2 5 4 3 1

Sample Output:

YESNONONOYESNO

#include<iostream>  #include<cstdio>  #include<cstring>  #include<cmath>  #include<algorithm>  #include<queue>  #include<stack>  #include<set>using namespace std;int main(){int n,m;int map[300][300]={0};scanf("%d%d",&n,&m);for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);map[a][b]=map[b][a]=1;}int k;scanf("%d",&k);while(k--){int len;int a[1000];scanf("%d",&len);int v[1000]={0};for(int i=0;i<len;i++){scanf("%d",&a[i]);v[a[i]]++;}if(a[0]!=a[len-1]||len!=n+1){printf("NO\n");continue;}int i;for(i=1;i<=n;i++){   if(v[i]!=1&&(i!=a[0]||i!=a[len-1])){break;    }    }    if(i!=n+1) printf("NO\n");else{int pro=a[0];int flag=0;for(int i=1;i<len;i++){if(map[a[i]][pro]){pro=a[i];}else{flag=1;break;}}if(flag) printf("NO\n");else printf("YES\n");} }    return 0;  }