PAT甲级1122 Hamiltonian Cycle (25 分)

1122 Hamiltonian Cycle (25 分)

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

      题目大意：

给出n个结点以及m条边，以及k个序列，每个序列都给出一个数num，以及num个结点序号，请问是否为汉密尔顿环，即是否为连通分量的一条路径。

思路：

         用邻接矩阵存储图，当遍历到e[a][b] == 0 时置flag = false，并且break出来，否则遍历完序列,如果flag == false那么输出NO就可以了，否则呢还要验证是否全部遍历过，用vis[id]来记录，且vis[start] == 2， 如果都是了才能输出YES。

         注意：如果num ！= n + 1时一定不是正确序列，这个可以增加判断，提高效率，此题中如果不进行判断的话，也能通过。

参考代码：

 

 

#include
#include
#include
using namespace std;
vector path, vis;
int e[201][201];
int n, m, k, num;
int main(){
	scanf("%d%d", &n, &m);
	vis.resize(n + 1);
	for(int i = 0; i < m; ++i){
		int a, b;
		scanf("%d%d", &a, &b);
		e[a][b] = e[b][a] = 1;
	}
	scanf("%d", &k);
	for(int i = 0; i < k; ++i){
		scanf("%d", &num);
		path.resize(num);
		for(int j = 0; j < num; ++j)	scanf("%d", &path[j]);
		if(num != n + 1){
			printf("NO\n");
			continue;
		}
		bool flag = true;
		int cnt = 0;
		for(int j = 0; j < num - 1; ++j){
			int a = path[j], b= path[j + 1];
			if(e[a][b] == 0){
				flag = false;
				break;
			}
			vis[a]++; vis[b]++; cnt++;
		}
		if(cnt != n || vis[path[0]] != 2) flag = false;
		if(!flag)	printf("NO\n");
		else printf("YES\n");
		fill(vis.begin(), vis.end(), 0);
	}
	return 0;
}