Uva524 Primer Ring Problem

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers $1, 2,……, 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 <= 16)

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.

You are to write a program that completes above process.

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

題解

找出n<=16的素數環，簡單的枚舉會超時，用回溯法

代碼實現

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 50
using namespace std;

int n, A[MAXN] = {1}, ispe[MAXN], vis[MAXN];

void dfs(int cur) //dfs回溯
{
    if(cur == n&& ispe[A[0] + A[n - 1]])
        {
            int k;
        for(int i = 1; i < n; i++)//輸出語句的控制
        {
            printf("%d ", A[i-1]);
            k=i;//最後一個i存儲在k中
        }
        printf("%d",A[k]);
        printf("\n");
    }
    else for(int i = 2; i <= n; i++)
    {
        if(!vis[i]&& ispe[i + A[cur - 1]])
        {
            A[cur] = i;
            vis[i] = 1;   //對其進行標記
            dfs(cur + 1);
            vis[i] = 0;     //清除標記
        }
    }
}

int main()
{
    for(int i = 2; i <= 50; i++)
        ispe[i] = 1;
    for(int i = 2; i <= 50; i++)
        for(int j = i + i; j + i <= 50; j += i)
            ispe[j] = 0;
    int kase = 0;
    while(cin >> n)
    {
        if(kase++)
            printf("\n");
        printf("Case %d:\n", kase);
        dfs(1);
    }
    return 0;
}