2016ACM暑假集训 - Prime Ring 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

解题思路：深度优先搜索，固定第一个为1，后面的为2-n，如果不符合条件救回溯。另外用一个数组储存是否是素数，加快速度

代码：

#include 
#include 
#include 
int a[25];
bool use[25];
int n;

bool p[] = {0, 0, 1, 1, 0, 1,
			0, 1, 0, 0, 0,
			1, 0, 1, 0, 0,
			0, 1, 0, 1, 0,
			0, 0, 1, 0, 0,
			0, 0, 0, 1, 0,
			1, 0, 0, 0, 0,
			0, 1, 0};

void DFS(int num)
{
    if(n == num && p[1+a[n-1]])
	{
        for(int i = 0; i < n; i++)
		{
            printf(i == n-1 ? "%d\n" : "%d ", a[i]);
        } 
    }
    else
	{
        for(int i = 2; i <= n; i++)
		{
            if(!use[i] && p[i+a[num-1]])
			{
                a[num] = i;
                use[i] = true;
                DFS(num+1);
                use[i] = false; 
            } 
        }
    }
}

int main()
{
    int t = 0;
    while(scanf("%d",&n) != EOF)
	{
        memset(use, 0, sizeof(use));
        a[0] = 1;
        t++;
        printf("Case %d:\n", t);
        DFS(1);
        puts("");
    }
    return 0;
}