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), 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 Kwhich 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

就多判断几下了，感觉都不涉及图的知识

#include <bits/stdc++.h>
using namespace std;
int n, m, k, p;
vector<int> vt;
int v[500][500];
int main(){
    cin >> n >> m;
    int x, y;
    for(int i = 0; i < m; i++){
        cin >> x >> y;
        v[x][y] = 1;
        v[y][x] = 1;
    }
    cin >> k;
    while(k--){
        cin >> p;
        vt.clear();
        for(int i = 0; i < p; i++){
            cin >> x;
            vt.push_back(x);
        }
        set<int> s;
        if(p == n+1){
            if(vt[0] == vt[vt.size()-1]){
                bool flag = true;
                for(int i = 1; i < vt.size(); i++){
                    if(v[vt[i-1]][vt[i]] == 0){
                        flag = false;
                        break;
                    }
                    s.insert(vt[i]);
                }
                if(flag && s.size() == n){
                    cout << "YES" << endl;
                }else{
                    cout <<"NO"<<endl;
                }
            }else{
                cout << "NO" << endl;
            }
        }else{
            cout << "NO" << endl;
        }
    }

    return 0;
}