HDU 1016 素数环-优快云博客

本文链接：https://blog.youkuaiyun.com/liujunhua_C/article/details/47440317

本文探讨了PrimeRingProblem的解决方法，这是一个经典的回溯法问题。博客详细介绍了如何使用回溯法找到所有可能的环形排列，使得相邻数字之和为质数，并提供了完整的C语言实现代码。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34394 Accepted Submission(s): 15216

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

/**
在朋友和学长的帮助下总算解决了这道题，都说是回溯法，可我依然
分不清深搜和回溯的差别，此题有待进一步分析。待续。。。。。。。
*/
#include<stdio.h>
#include<string.h>

int n;
int vis[30];
int a[30];
int count = 0;

bool is_prime(int n)
{
    for(int i = 2; i*i <= n; i++)
        if(n%i == 0)
            return false;
    return true;
}

void dfs(int cur)
{
    if(cur == n && is_prime(a[0]+a[n-1]))///递归的边界条件，当最后一个位置放好的时候递归结束，由于是环，需判断首尾是否满足条件
    {
        printf("%d",a[0]);
        for(int i = 1; i < n; i++)
            printf(" %d",a[i]);
        printf("\n");
    }
    else for(int i = 2; i <= n; i++)///所有的可放元素
        {
            if(!vis[i] && is_prime(i+a[cur-1]))///判断当前元素是否满足条件
            {
                a[cur] = i;
                vis[i] = 1;///用过的元素标记一下
                dfs(cur+1);
                vis[i] = 0;///回溯的时候还原现场
            }
        }
}

int main()
{
    while(~scanf("%d",&n))
    {
        if(n%2!=0)
        {
            printf("Case %d:\n\n",++count);
            continue;
        }
        else
            printf("Case %d:\n",++count);
        a[0] = 1;
        memset(vis,0,sizeof(vis));
        dfs(1);
        printf("\n");
    }
    return 0;
}