1122. Hamiltonian Cycle (25)

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本文介绍了一个算法问题，即判断给定的路径是否构成Hamiltonian Cycle。输入包括顶点数、边数、图的边和查询路径，输出为YES或NO，指示路径是否形成Hamiltonian Cycle。

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时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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

C++：

/*
 @Date    : 2018-03-02 09:44:15
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1122
 */

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

static int N, M;
static const int MAXN = 201;
static bool G[MAXN][MAXN];
static bool checked[MAXN];

int main(int argc, char const *argv[])
{
    scanf("%d %d", &N, &M);
    int v, w;
    for (int i = 0; i != N; ++i)
        for (int j = 0; j != N; ++j)
            G[i][j] = false;
    while (M--)
    {
        scanf("%d %d", &v, &w);
        G[v-1][w-1] = G[w-1][v-1] = true;
    }
    int K;
    scanf("%d", &K);
    while (K--)
    {
        int n;
        scanf("%d", &n);
        int path[n];
        for (int i = 0; i != n; ++i)
            scanf("%d", &path[i]);
        bool ok = false;
        if (n == N + 1 && path[N] == path[0])
        {
            ok = true;
            fill(checked, checked+N, false);
            for (int i = 0; i != N; ++i)
            {
                if (!G[path[i]-1][path[i+1]-1] || checked[path[i]-1])
                {
                    ok = false;
                    break;
                }
                checked[path[i]-1] = true;
            }
        }
        if (ok) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}