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[][];
 int path[],vis[];
 int main(){
     scanf("%d %d",&n,&m);
     for(int i=;i<m;i++){
         int c1,c2;
         scanf("%d %d",&c1,&c2);
         dis[c1][c2]=;
         dis[c2][c1]=;
     }
     scanf("%d",&k);
     int min=,mini=;
     for(int i=;i<=k;i++){
         int flag=;
         int total=;
         int nn;
         fill(vis,vis+,);
         scanf("%d",&nn);
         for(int j=;j<nn;j++){
              scanf("%d",&path[j]);
              vis[path[j]]++;
         }
         for(int j=;j<=n;j++){
              if(vis[j]==) flag=;
         }
         if(nn!=n+ || path[]!=path[nn-]) flag=;
         for(int j=;j<nn;j++){
               if(dis[path[j]][path[j-]]==){
                   flag=;
                   break;
             }
           }
           if(flag==) printf("NO\n");
           else{
               printf("YES\n");
         }
     }
 }

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