DFS专题 Prime Ring Problem

# 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