A1122. Hamiltonian Cycle

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 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<iostream>
 #include<vector>
 using namespace std;
 const int INF = ;
 int G[][], visit[] = {,};
 int main(){
     int N, M;
     fill(G[], G[] + *, INF);
     scanf("%d%d", &N, &M);
     for(int i = ; i < M; i++){
         int v1, v2;
         scanf("%d%d", &v1, &v2);
         G[v1][v2] = G[v2][v1] = ;
     }
     int K;
     scanf("%d", &K);
     for(int i = ; i < K; i++){
         int n, s, vi, vj, tag = ;
         fill(visit, visit + , );
         scanf("%d%d", &n, &s);
         if(n != N + )
             tag = ;
         visit[s] = ;
         vi = s;
         for(int j = ; j < n; j++){
             scanf("%d", &vj);
             if(G[vi][vj] == INF){
                 tag = ;
             }
             visit[vj]++;
             vi = vj;
         }
         if(vj != s)
             tag = ;
         for(int i = ; i <= N; i++){
             if(i != s && visit[i] !=  || i == s && visit[i] != )
                 tag = ;
         }
         if(tag == )
             printf("NO\n");
         else printf("YES\n");
     }
     return ;

 }

总结：

1、哈密顿回路：图中所有顶点必须都出现，除了首尾是重复出现外，其它节点仅出现一次。   顶点组成的序列必须是联通的。

2、注意，边读入边做处理时，不要使用break，造成读入数据错乱。