HDU 1016 Prime Ring Problem【深搜练习】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55230 Accepted Submission(s): 24442

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

题意概括：

　　对于数字1~n，对其进行排序，要求是当前数和其后一个数之和必须是素数，并且1永远都在第一位并和最后一个数之和也为素数。

解题分析：

　　其实就是用深搜对1~n的数列进行全排列，输出其中特定的序列。

AC代码：

#include<stdio.h>#include<string.h>#include<math.h>#define N 23int book[N], a[N] = {1}, n;int IsPrime(int m)//因为数据较小，也可以用一个数组来记录是否是素数{    if(m == 1) return 0;    int i, k = (int)sqrt(m);    for(i = 2; i <= k; i++)        if(m%i == 0) return 0;    return 1;}void dfs(int step){    int i;    if(step >= n){        if(IsPrime(1+a[step-1])){            printf("1");            for(int j = 1; j < n; j++)                printf(" %d", a[j]);            printf("\n");        }        return ;    }    for(i = 2; i <= n; i++){        if(!book[i] && IsPrime(a[step-1]+i)){            book[i] = 1;            a[step] = i;            dfs(step+1);            book[i] = 0;        }    }}int main(){    int i, t = 0;    while(~scanf("%d", &n)){        memset(book, 0, sizeof(book));        printf("Case %d:\n", ++t);        dfs(1);        printf("\n");    }    return 0;}