PAT-甲级-1122 Hamiltonian Cycle (25 分)

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本文详细介绍了一种用于判断图中是否存在Hamiltonian Cycle的算法，包括输入输出规范、样例及解题思路。通过检查路径点数、起点与终点是否一致、所有顶点是否唯一出现一次且每对相邻顶点间存在边，来判断给定路径是否构成Hamiltonian Cycle。

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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 V1 V2 ... Vn

where n is the number of vertices in the list, and Vi'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

解题思路：

查询点数p = n+1;
起点s = 终点d；
！！相邻点有连接边；
无重复点数 = n；

#include <iostream>
#include <set>
using namespace std;
#define rep(i,j,k)    for(int i=j;i<k;i++)
int v[210][210];
int main(){
    std::ios::sync_with_stdio(false);
    int n,m,k,a,b,p;
    cin>>n>>m;
    fill(v[0],v[0]+210*210,-1);
    rep(i,0,m){
        cin>>a>>b;
        v[a][b]=v[b][a]=1;
    }
    cin>>k;
    rep(i,0,k){
        int s,d,flag=1,pre;
        set<int> st;
        cin>>p>>pre;
        s = pre;
        rep(j,1,p){
            cin>>a;
            if(v[pre][a]!=1) flag=0;
            if(j==p-1)    d=a;
            st.insert(a);
            pre = a;
        }
        if(st.size()!=n || p != n+1 || s!=d || flag==0){
            cout<<"NO";
        }else cout<<"YES";
        if(i!=k-1)    cout<<"\n";
    }
    return 0;
}