PAT - 甲级 - 1122. Hamiltonian Cycle (25)

题目描述：

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

题目思路：

题目中只是说了一个环中，包含图中所有的元素，那么这样的环就是Hamiltonian环。题目中的描述给出的信息是非常有限的，当然如果你知道哈密顿回路的定义，那么你只需要稍微一看就知道题目要求什么了。言归正传，因为题目中给定的信息有限，我们需要根据给出的测试用例来推断题目的要求。根据题目给出的用例，我们可以推断这个环具备的条件（属性）。

1.包含所有元素

2.第一个元素和最后一个元素相同（即能构成环）

3.元素个数一定不小于n

但是还有一个疑问，元素是否可以重复？（除第一个和最后一个元素）。

根据题目给定的描述是无法判定的，我们可以先提交一次不判重的代码（或者判重的代码）。果然有一个测试用例的答案是错误的，因此元素是不能重复的。

题目代码：

#include
#include
#include
using namespace std;
int e[205][205];
int n, m, t, k, a[205], x, y,flag = 1;
int main(){
	scanf("%d%d",&n,&m);
	memset(e,0,sizeof(e));
	// 数据输入 
	for(int i = 0; i < m; i++){
		scanf("%d%d",&x, &y);
		e[x][y] = e[y][x] = 1;
	}
	scanf("%d",&t);
	while(t--){
		flag = 1;
		scanf("%d",&k);
		 
		if(k != n+1){ // 判断元素个数 
			for(int i = 0; i < k; i++){
				scanf("%d",&a[0]);
			}
			flag = 0;
		}else{
			for(int i = 0; i < k; i++){
				scanf("%d", &a[i]);
			}
			if(a[0] != a[k-1]){ // 首尾元素比较 
					flag = 0;
			}else{
				for(int i = 0; i < k-1; i++){ // 是否通路判断 
					if(!e[a[i]][a[i+1]]){
						flag = 0;
						break;
					}
				}
				sort(a,a+n-1);
				for(int i = 0; i < k-1; i++){ // 判断重复元素 
					if(a[i] == a[i+1]){
						flag = 0;
						break;
					}
				}
			}
		}
		if(flag)puts("YES");
		else puts("NO");
	}	
	return 0;
}