hdoj 1016 素数圆环问题（深度优先搜索z和道题 ）

        这道题出现的最大问题是超时，后来参照网上的代码才明白原因：

  按照自己的思想，先判断数字是否重复，此过程经历一系列递归，然后又在这些不重复的序列循环搜索素数序列，这样就经历了两回循环，必然耗时，而进行优化后，可以在递归过程用两个条件判断，既要求数字不重复，又要求相邻和两端必须是素数！大大减少了运行时间！

                                 Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 24   Accepted Submission(s) : 7

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

     

       

                                       

AC码

#include <cstdio>
#include <cstring>
int n,m,a[21],arr[21]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20},mark[21];
int c[50];
int k=1,j;
void dfs(int v)
{
	 if(v==n&&c[a[0]+a[n-1]])  
     {  
              for(int r=0;r<n;r++)  
               {  
                      if(!r)printf("%d",a[r]);  
                      else printf(" %d",a[r]);  
               }  
              printf("\n"); 
			  return;
	 }
    /*if(v >= n){
    	for(int i=1;i<n;i++)     //这种方法超时 
    	{
    		if(!c[a[i]+a[i-1]])			
    		return;
    		if(!c[a[0]+a[n-1]])        
    		return;  
		}
		if(a[0]==1)  
		{
			for(int i = 0;i < n;i++)
           {	
           		
				   printf("%d ",a[i]);
		   } 
        printf("\n");
        return ;
		}*/
        

    for(int j = 1; j< n;j++)
	{
        if(!mark[j])   //打印出来的数字不能重复 
		{
        	if(c[a[v-1]+arr[j]]) //同时判断必须是素数 
            {
            	mark[j] = 1;  //标记为已用过 
            	a[v] = arr[j]; 
            	dfs(v+1);
            	mark[j] = 0;  //如果退回来一步就必须把原位置标记为没用过 
			}
			else continue;
        }
    }
    
}
int main()
{
    while(scanf("%d",&n)==1)
	{
        memset(mark,0,sizeof(mark));
        c[2]=c[3]=c[5]=c[7]=c[11]=c[13]=c[17]=c[19]=c[23]=c[29]=c[31]=c[37]=c[39]=1; 
		printf("Case %d:\n",k++) ;
		a[0]=1;//因为a[0]已经确定，直接写出来可以节省判断的时间 
        dfs(1);   //a[0]已经打印，从第二位开始判断 
        printf("\n");
    }
    
}