Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25700    Accepted Submission(s): 11453

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

题意：n个人围在一起，任意2个相邻的之和为素数，输出所有的可能的情况，但情况还是要按一定的规则输出，就是前面的数字由小到大的输出情况。每个case后右一个空格，注意一下格式吧。

解题思路：从第一个位置开始，判断下一个位置加入哪一个数，并判断是否合适。若合适则标记加入，不断for循环，做好DFS的判断条件。

贴出代码：

#include <stdio.h>
#include <string.h>

int visited[25], foot[25];
int x[11]= {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

int Prime(int a, int b)
{
    int  mark = 0;

    for(int i = 0; i<11; i++)
    {
        if( a+b == x[i] )
        {
            mark = 1;
            break;
        }
    }

    return mark;
}

void DFS(int n, int m)
{
    if(m == n)
    {
        if(Prime(foot[1], foot[n]))
        {
            for(int i = 1; i<n; i++)
            {
                printf("%d ", foot[i]);
            }
            printf("%d\n", foot[n]);
            //printf("\n");
        }
    }
    else
    {
        for(int i = 1; i<=n; i++)
        {
            if(visited[i])
                continue;
            if(Prime(foot[m], i) == 1)
            {
                foot[m+1] = i;
                visited[i] = 1;
                DFS(n, m+1);
                visited[i] = 0;
            }
        }
    }
}

int main()
{
    int n, k = 1;
    while(scanf("%d", &n)!=EOF)
    {
        if(n<=0 || n>=20)
            break;
        memset(visited, 0, sizeof(visited));
        visited[1] = 1;
        foot[1] = 1;
        printf("Case %d:\n", k++);
            DFS(n, 1);
        printf("\n");
    }

    return 0;
}

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