PAT--1122. Hamiltonian Cycle (25)

Description

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

题解

给定一个无向图，以及遍历序列，判断其是否为哈密顿回路。

考虑哈密顿回路的必要条件：

• 序列首尾相同
• 序列长度 = 顶点个数 + 1
• 除首尾外，不会再有相同的顶点。
• 序列中相邻的两个顶点要求有边相连。

满足这些条件可以判断这是一条哈密顿回路。

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <queue>#include <map>#include <string>#include <set>using namespace std;const int maxn = 200 + 10;int n, m, k;vector<int> G[maxn];bool connect(int u, int v){    for(int i = 0; i < G[u].size(); ++i){        if(G[u][i] == v) return true;    }    return false;}bool judge(vector<int>& path){    int len = path.size();    if(len != n + 1) return false;    if(path[0] != path[len - 1]) return false;    vector<int> vis(path.begin(), path.end() - 1);    sort(vis.begin(), vis.end());    for(int i = 0; i < vis.size(); ++i){        if(vis[i] != i + 1) return false;    }    for(int i = 0; i < path.size() - 1; ++i){        int u = path[i], v = path[i + 1];        if(!connect(u, v)) return false;    }    return true;}int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif    cin >> n >> m;    int u, v;    for(int i = 0; i < m; ++i){        cin >> u >> v;        G[u].push_back(v);        G[v].push_back(u);    }    cin >> k;    for(int i = 0; i < k; ++i){        int t; cin >> t;        vector<int> path;        path.resize(t);        for(int j = 0; j < t; ++j){            cin >> path[j];        }        if(judge(path)) cout << "YES" << endl;        else cout << "NO" << endl;    }    return 0;}