HDU 1016 - Prime Ring Problem

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

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10426    Accepted Submission(s): 4690

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

 

很经典的dfs加回溯，在白书上看过，所以很快就ac了。 需要注意不要忘了判断最后一个数是否与第一个数的和为素数，因为是素数环~~~呵呵~~

代码如下：

#include<iostream>
#include<math.h>
#include<cstring>
using namespace std;
int prime[40];
int d[40];
int vis[40];
int is_prime(int x)
{
    if(x <= 1)
        return 0;
    for(int i = 2; i <= sqrt(x); i++)
        if(x % i == 0)
            return 0;
    return 1;
}
void dfs(int cur, int n)
{
    if(cur == n)
    {
        if(prime[d[0] + d[n-1]])
        {
            for(int i = 0; i < n; i++)
            {
                printf("%d", d[i]);
                if(i != n-1)
                    putchar(' ');
            }
            putchar(10);
        }
        return;
    }
    for(int i = 2; i <= n; i++)
    {
        if(!vis[i] && prime[i + d[cur-1]])
        {
            d[cur] = i;
            vis[i] = 1;
            dfs(cur + 1, n);
            vis[i] = 0;
        }
    }
}
int main()
{
    int n, cnum = 0;
    for(int i = 0; i < 40; i++)
        prime[i] = is_prime(i);
    while(scanf("%d", &n) != EOF)
    {
        cnum++;
        memset(vis, 0, sizeof(vis));
        printf("Case %d:\n", cnum);
        d[0] = 1;
        dfs(1, n);
        putchar(10);
    }
    return 0;
}