hdu-1016 Prime Ring Problem 搜索题

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

简单的搜索题目，不过我还是好长时间没通过。。。
code:
#include<stdio.h>
#include<string.h>
int num[40];
int mark[120];
int prime[40];
void dfs(int root,int n,int m){
	int i,j;
	if(m>n&&!prime[num[n]+num[1]]){     //当搜索完后且第一个数和最后一个数也符合条件 
		for(j=1;j<n;j++)                //的时候输出 
		   printf("%d ",num[j]);
		printf("%d\n",num[n]);          //前边的数跟后边的数分开输出 
		return;
	}
	for(i=1;i<=n;i++){
		if(!prime[i+root]&&!mark[i]){     //对数进行判断，如果每相邻两个数之和都为素数且未被标记 
			num[m]=i;                     //则满足条件 
			mark[i]=1;                    //对处理过的数字进行标记 
			dfs(i,n,m+1);                 //返回下一个要搜索的数字 
			mark[i]=0;
		}
	}
}
int main(){
	int k,n,m,t,i,j;
	int cas=1;
	memset(prime,0,sizeof(prime));
	prime[0]=prime[1]=1;
	for(i=2;i*i<40;i++){     //打表，不是素数的标记为1 
		if(!prime[i])
		   for(j=i*i;j<40;j+=i)
		      prime[j]=1;
	}

	while(~scanf("%d",&n)){
		memset(mark,0,sizeof(mark));
		printf("Case %d:\n",cas++);
		num[1]=1;
		mark[1]=1;//1不用判断，直接标记 
		dfs(1,n,2);
		printf("\n");
	}
	return 0;
} 


南阳oj上边有一道几乎相同的题目：
题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=488
大体思路还是相同的

代码：
#include<stdio.h>
#include<string.h>
int mark[50];
int num[50];
int prime[50];
void dfs(int root,int n,int m){
	int i,j;
	if(m>n&&!prime[num[1]+num[n]]){
		for(j=1;j<=n;j++)
		   printf("%d ",num[j]);
		printf("\n");
	//	printf("%d\n",num[n]);
	}
	 // printf("No Answer\n");
	for(i=1;i<=n;i++){
		if(!mark[i]&&!prime[i+root])
		{
			num[m]=i;
			mark[i]=1;
			dfs(i,n,m+1);
			mark[i]=0;
		}
	}
	return;
}
int main(){
	int i,j,k,t;
	int n,m;
	int Case=1;
	memset(prime,0,sizeof(prime));
	prime[0]=prime[1]=1; 
	for(i=2;i<50;i++){
		if(!prime[i])
		   for(j=i*i;j<50;j+=i)
		     prime[j]=1;
	}
	//for(i=0;i<50;i++){
	//	if(!prime[i])
	//	  printf("%d ",i);
	//}
	while(~scanf("%d",&n),n!=0){
		memset(mark,0,sizeof(mark));
		memset(num,0,sizeof(num));
		num[1]=1;
		mark[1]=1;
		printf("Case %d:\n",Case++);
		if(n==1)
		   printf("%d\n",1);//这部分比较容易忽略， 
		else if(n&1)
		    printf("No Answer\n");
		else 
		dfs(1,n,2);
		
	//	printf("\n");
	}
	return 0;
}