HDU 1016 Prime Ring Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1016

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
 
时间复杂度:$O(n!)$
代码:
#include <stdio.h>
#include <string.h>
int book[30], a[30], n;
int prime(int x) {
    int i;
    for (i = 2; i <= x / 2; i ++)
        if (x % i == 0) return 0;
    return 1;
}

void dfs(int step) {
    if (step == n + 1 && prime(a[1] + a[n])) {
        for (int i = 1; i <= n - 1; i ++)
            printf("%d ", a[i]);
        printf("%d
", a[n]);
        return;
    }
    
    for (int i = 2; i <= n; i ++) {
        if (book[i] == 0) {
            if (prime(i + a[step - 1])) {
                book[i] = 1;
                a[step] = i;
                dfs(step + 1);
                book[i] = 0;
            }
        }
    }
    return;
}

int main() {
    int kase = 0;
    a[1] = 1;
    while (~scanf("%d", &n)) {
        memset(book, 0, sizeof(book));
        printf("Case %d:
", ++kase);
        dfs(2);
        printf("
");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/zlrrrr/p/9465172.html