HDU1016：Prime Ring Problem（dfs）

最新推荐文章于 2019-01-29 21:58:59 发布

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本文介绍了一道经典的算法题目——素数环问题。任务要求利用1到n的数字构成一个环，使得任意相邻两个数字之和为素数，并按字典序输出所有可能的组合方案。文中提供了一个C++实现示例，采用深度优先搜索算法完成求解。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52756 Accepted Submission(s): 23320

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

Input

n (0 < n < 20).

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

Sample Input

6
8

Sample Output

  
   Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

Source

Asia 1996, Shanghai (Mainland China)

题意：给n个数（1~n），要求组成一个环，要求相邻两个元素之和是素数，字典序输出所有组合。

心情差就想做水题。

# include <iostream>
# include <cstdio>
# include <cstring>
using namespace std;
int n, a[23], p[45], vis[23];
void dfs(int pos)
{
    if(pos == n+1)
    {
        for(int i=1; i<=n; ++i)
            printf("%d%c",a[i],i==n?'\n':' ');
        return;
    }
    for(int i=2; i<=n; ++i)
    {
        if(vis[i]||p[i+a[pos-1]]) continue;
        if(pos == n && p[i+a[1]]) continue;
        a[pos] = i;
        vis[i] = 1;
        dfs(pos+1);
        vis[i]=0;
    }
}
int main()
{
    int cas=1;
    for(int i=2; i<=40; ++i)
        if(p[i] == 0)
            for(int j=i+i; j<=40; j+=i)
                p[j] = 1;
    while(~scanf("%d",&n))
    {
        a[1] = 1;
        printf("Case %d:\n",cas++);
        dfs(2);
        puts("");
    }
    return 0;
}