hdu 1016 DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57319    Accepted Submission(s): 25008

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.

 

题意：给你一个1到n个点的环，1必须在第一位,要相邻两点相加为素数。输出所有情况。

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

代码：

 

#include<stdio.h>
#include<string.h>
int visit[30],a[30],n;//a为第几个数 
int su(int x)
{
    for(int i=2;i<x;i++)
        if(x%i==0) return 0;
    return 1;
}
void dfs(int x,int y)//x为深度，也是第几位数，y为1~n中的值 
{
    a[x]=y;
    visit[y]=1;
    if(x==n)//终止条件 
    {
        if(su(a[x]+a[1]))//还要判断一下最后一个与1是否构成素数 
        {
            printf("1");
            for(int i=2;i<=n;i++)
                printf(" %d",a[i]);
            printf("\n");
        }
        return ;
    }
    for(int i=1;i<=n;i++)
    {
        if(!visit[i]&&su(a[x]+i))
        {
            dfs(x+1,i);//如果i与a[x]构成素数，搜索下一层 
            visit[i]=0;//回溯 
        }
    }
    return ;
}
int main()
{
    int sum=1;
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case %d:\n",sum++);
        memset(a,0,sizeof(a));
        dfs(1,1);
        printf("\n");
    }
    return 0;
}