PAT 1122 Hamiltonian Cycle

Posted 2021-11-27 cunyusup

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 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

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 500

int n,m;
int G[MAXN][MAXN] = {0};





int main(){
    cin >> n >> m;
    for(int i=0;i < m;i++){
        int x,y; cin >> x >> y;
        G[x][y] = G[y][x] = 1;
    }

    int K;
    cin >> K;
    while(K--){
        int nnn; cin >> nnn;
        vector<int> temp;
        set<int> st;
        for(int i=0;i < nnn;i++){
            int num;cin >> num;
            temp.push_back(num);
            st.insert(num);
        }
        int flag1=1;
        int flag2=1;
        if(st.size()!=n||nnn-1!=n||temp[0]!=temp[nnn-1]) flag1=0;

        for(int i=1;i < temp.size();i++){
            if(!G[temp[i-1]][temp[i]])flag2=0;
        }
        if(flag1&&flag2) cout << "YES" << endl;
        else cout << "NO" << endl;

    }




    return 0;
}

看了答案才知道这么简单，一开始理解错了题意，后来才发现他给的就是路径，难度就降低了

分析：1.设置falg1 判断节点是否多走、少走、或走成环
2.设置flag2 判断这条路能不能走通
3.当falg1、flag2都为1时是哈密尔顿路径，否则不是