DFS——hdu1016Prime Ring Problem

一、题目回顾

题目链接：Prime Ring Problem

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

 

题意：输入一个数n,把1到n的自然数放到一个环里，保证相邻的两个数的和是素数。

 

 

二、解题思想

• dfs+素数打表
• 经典的一道DFS

三、代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 50;
int a[maxn];
int b[maxn];
int n,prime[2*maxn];
bool vis[maxn];

//素数打表，相当于int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};
void isprime()
{
    int i,j;
    for(i = 0;i<50;i++)
    prime[i] = 1;
    prime[0] = prime[1] = 0;
    for(i = 2;i<50;i++)
    {
        if(prime[i])
        for(j = i+i;j<50;j+=i)
        prime[j] = 0;
    }
}
//深搜
void dfs(int x)									//x：当前搜索第几个数
{
	if(x==n+1 && prime[b[n]+b[1]]){				//满足条件了，就输出来
		for(int i=1;i<n;i++)	printf("%d ",b[i]);
		printf("%d\n",b[n]);
		return;
	}
	for(int i=2;i<=n;i++){
		if(!vis[i] && prime[a[i]+b[x-1]]){		//此数未用并且与上一个放到环中的数相加是素数
			vis[i] = 1;							//标记
			b[x] = a [i];						//放进数组
			dfs(x+1);
			vis[i] = 0;							//退去标记
		}
	}

}
int main()
{
	int kase = 1;
	isprime();
	while(cin>>n && (n>0&&n<20)){
		for(int i=1;i<=n;i++)
			a[i] = i;
		memset(vis,0,sizeof(vis));
		b[1] = 1;
		printf("Case %d:\n",kase++);
		dfs(2);
		printf("\n");
	}
	return 0;
}