PAT A1122. Hamiltonian Cycle (25)

1. 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 … Vn

where n is the number of vertices in the list, and Vi’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

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=202;
int G[maxn][maxn];
int vis[maxn];
int temp[maxn];
int n,m,q,k;
bool flag;
int main(){
    fill(G[0],G[0]+maxn*maxn,0);
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d %d",&a,&b);
        G[a][b]=G[b][a]=1;
    }
    scanf("%d",&q);
    for(int i=0;i<q;i++){
        fill(vis,vis+maxn,0);
        flag=true;
        scanf("%d",&k);
        for(int j=0;j<k;j++){
            scanf("%d",temp+j);
            vis[temp[j]]=1;
        }
        if(k!=n+1){
            flag=false;
        }
        if(temp[0]!=temp[k-1]){
            flag=false;
        }
        for(int j=0;j<k-1;j++){
            if(G[temp[j]][temp[j+1]]!=1){
                flag=false;
            }
        }
        for(int j=1;j<=n;j++){
            if(vis[j]==0){
                flag=false;
            }
        }
        if(flag==true){
            printf("YES\n");
        }else printf("NO\n");
    }
    return 0;
}