PAT A1122 Hamiltonian Cycle （25 分）——图遍历

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

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <set>
using namespace std;
int n,m,k;
int dis[201][201];
int path[1000],vis[201];
int main(){
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;i++){
        int c1,c2;
        scanf("%d %d",&c1,&c2);
        dis[c1][c2]=1;
        dis[c2][c1]=1;
    }
    scanf("%d",&k);
    int min=99999999,mini=0;
    for(int i=1;i<=k;i++){
        int flag=0;
        int total=0;
        int nn;
        fill(vis,vis+201,0);
        scanf("%d",&nn);
        for(int j=0;j<nn;j++){
             scanf("%d",&path[j]);
             vis[path[j]]++;
        }
        for(int j=1;j<=n;j++){
             if(vis[j]==0) flag=1;
        }
        if(nn!=n+1 || path[0]!=path[nn-1]) flag=1;
        for(int j=1;j<nn;j++){
              if(dis[path[j]][path[j-1]]==0){
                  flag=1;
                  break;
            }
          }
          if(flag==1) printf("NO
");
          else{
              printf("YES
");
        }
    }
}

注意点：挺简单的一道题，就按题目意思实现一下，判断给的路径是否是一个环，这个环有没有包含所有节点，并且是联通的，点除了起点都只经过一次。和1150 Travelling Salesman Problem （25 分）差不多，比他更简单一些