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：给出一个图，判断给定的路径是不是哈密尔顿路径

分析：

1.如果节点多走、少走、或走成环，则输出NO
2.如果这条路不能走通，则输出NO.
3.最后剩下的就一定是哈密尔顿路径，输出YES。

#include <iostream>
#include <vector>
#include <set>
using namespace std;
int e[300][300], n, m;
vector<int> v;
void check(int index) {
	    int k;
        cin >> k;
        vector<int> v(k);
        set<int> s; 
        for(int i = 0; i < k; i++) {
            scanf("%d", &v[i]);
            s.insert(v[i]);
        }
        if( s.size() != n || k - 1 != n || v[0] != v[k-1] ){ 
            printf("NO\n");
            return ;
		}
	    for(int i = 0; i < k - 1; i++)
            if( e[v[i]][v[i+1]] == 0) {
            	printf("NO\n");
                return ;
			}
       printf("YES\n");
}
int main() {
    scanf("%d%d", &n, &m);
    for ( int i = 0; i < m; i++) {
          int t1, t2, t;
          scanf("%d%d", &t1, &t2);
          e[t1][t2] = e[t2][t1] =1;
    }
    int k;
    scanf("%d", &k);
    for ( int i = 1; i <= k; i++) 
	      check(i); 
    return 0;
}