[HDU1016]Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 42782    Accepted Submission(s): 18977

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

 

题解：题目的意思很明确，首先，开头要为1，然后相邻两个相加素数，头尾相加也要为素数（一个函数搞定）此题类似八皇后，简单dfs，

#include 
#include 
#include 
using namespace std;
const int maxn = 10001;
int d[maxn];
int jd[maxn];//用来标记我这个数是否已经取过了
int n;
int count;
void dfs(int depth);
bool check(int n)
{
	int i;
	for(i = 2;i*i<=n;++ i)
		if(n%i==0)
			return false;
	return true;
}
void dfs(int depth)
{
	if(depth>n){//直到我走到n+1，说明我已经填完了
		if(check(d[1]+d[n])){//判断头尾是否相加为素数
			cout << d[1];
			for(int j = 2;j <= n;++ j)
				cout << ' ' << d[j];
			cout << endl;
		}
	}
	for(int x = 2;x <= n;++ x)
	{
		if(check(d[depth-1]+x)&&!jd[x]){
			d[depth] = x;
			jd[x]=1;
			dfs(depth+1);
			jd[x]=0;
		}
	}
}
int main()
{
	while(scanf("%d",&n)==1)
	{
		printf("Case %d:\n",++count);
		memset(d,0,sizeof(d));
		memset(jd,0,sizeof(jd));
		d[1]=jd[1]=1;
		dfs(2);
		cout << endl;
	}
}