DFS专题 Prime Ring Problem

View Code

# include <cstdio>

# include <cstring>



# define N 20 + 5



char ptab[25] = {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};



int n, solu[N];

bool vis[N];



void dfs(int cnt)

{

    if (cnt == n && ptab[solu[cnt]+1])

    {

        printf("1");

        for (int i = 2; i <= n; ++i)

            printf(" %d", solu[i]);

        putchar('\n');

        return ;

    }

    for (int i = 2; i <= n; ++i)

    {

        if (vis[i] == false && ptab[solu[cnt]+i])

        {

            vis[i] = true;

            solu[cnt+1] = i;

            dfs(cnt+1);

            vis[i] = false;

        }

    }

}



void solve(void)

{

    solu[1] = 1, vis[1] = true;

    memset(vis+1, false, sizeof(vis[0])*n);

    dfs(1);

    putchar('\n');

}



int main()

{

    int i = 0;



    while (~scanf("%d", &n))

    {

        printf("Case %d:\n", ++i);

        solve();

    }



    return 0;

}

 

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