HDU1016 Prime Ring Problem(素数环，深搜DFS)

题目：

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45061    Accepted Submission(s): 19969

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

思路：

利用搜索，枚举每一种可能，然后把符合条件的输出

代码：

#include <stdio.h>#include <string.h>#define mem(a,b) memset(a,b,sizeof(a))int prime[100]= {1,1};int a[50],md[50],n;//a来存放素数环里的数，md是做标记void sushu(){    int i,j;    for(i=2; i<=10; i++)        if(prime[i]==0)            for(j=2*i; j<=50; j+=i)                prime[j]=1;//把不是素数的标记为1}void dfs(int step){    int i,j;    if(step==n+1&&prime[a[n]+a[1]]==0)//结束条件，这个代码是从a[1]开始的    {        for(i=1; i<n; i++)            printf("%d ",a[i]);        printf("%d\n",a[n]);        return;    }    for(i=2; i<=n; i++)    {        if(md[i]==0&&prime[i+a[step-1]]==0)        {            a[step]=i;//给素数环填数            md[i]=1;//标记一下走过的            dfs(step+1);//搜索下一步            md[i]=0;//标记回来        }    }    return;}int main(){    int num=1;    a[1]=1;//从1开始的    sushu();    while(~scanf("%d",&n))    {        mem(md,0);        printf("Case %d:\n",num++);        dfs(2);        printf("\n");    }    return 0;}