hdu1016Prime Ring Problem 素数环

Prime Ring Problem

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

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

 

自己独立写的第一个dfs，很好很好。

之后发现小白上有原题。。。。压力很大。。

AC代码：

#include<iostream>
#include<cstring>
#define MAX 30
using namespace std;

int ring[MAX], vis[MAX], num, cnt = 1;
const int pri[11] = {3,5,7,11,13,17,19,23,29,31,37};

void dfs(int cur)
{
	if (cur == num - 1)
	{
//		cout << cur << ' ' << ring[num - 1] + 1 << endl;
		int ok = 0;
		for (int i = 0; i < 11; i++)
			if (ring[num -1] + 1 == pri[i])
				ok = 1;
		if (ok == 0)
			return;
		for (int i = 0; i < num - 1; i++)
			cout << ring[i] << ' ';
		cout << ring[num - 1] << endl;
		return;
	}
	vis[ring[cur]] = 1;
	for (int i = 0; i < 11; i++)
	{
		if (vis[pri[i] - ring[cur]] == 0 && pri[i] - ring[cur] > 1)
		{
		 	if (pri[i] - ring[cur] > num)
				break;
			ring[cur + 1] = pri[i] - ring[cur];
			dfs(cur + 1);
		}
	}
	vis[ring[cur]] = 0;
}

int main()
{
	while(cin >> num)
	{
		memset(ring, 0, sizeof (ring));
		memset(vis, 0, sizeof (vis));
		ring[0] = 1;
		cout << "Case " << cnt << ':' << endl;
		cnt++;
		dfs(0);
		cout << endl;
	}
}