PAT A1122 Hamiltonian Cycle (25point(s))

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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 ... Vn ​​ 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

题意

判断是否为哈密尔顿路径

解答

#include<iostream>
#include<vector>
#include<unordered_map>

using namespace std;

int main()
{
	int vn, en, a, b, testn, n;
	cin >> vn >> en;
	vector<unordered_map<int, bool>> adj(vn + 1);
	for (int i = 0; i < en; i++) {
		cin >> a >> b;
		adj[a][b] = true;
		adj[b][a] = true;
	}
	cin >> testn;
	for (int i = 0; i < testn; i++) {
		bool flag = true;
		cin >> n;
		unordered_map<int, bool> mp;
		int begin, pre;
		cin >> begin;
		mp[begin] = true;
		pre = begin;
		a = pre;
		for (int j = 1; j < n; j++) {
			cin >> a;
			mp[a] = true;
                        // 坑点,要判断是否连通
			if (adj[pre].find(a) == adj[pre].end()) 
				flag = false;
			pre = a;
		}
		if (flag && begin == a && mp.size() == vn && n == vn + 1) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}