杭电1016深度搜索问题

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14061    Accepted Submission(s): 6390

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

 

这道题可以通过深度搜索的办法来解决，下面是我的代码，已经通过了AC

//1016
#include <iostream>
#include <vector>
using namespace std;
int result[20];//存放计算的结果
vector<vector<int> > results;
int n;
//int primes[] = {2,3,5,7,11,13,17,19,23,29,31,37};//共有13个40以内的素数
bool prime[] = {0,0 ,1 ,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1};

bool isPrime(int num)
{
	return prime[num];
}

bool isIn(int list[], int len, int num)
{
	for (int i = 0; i < len; ++ i)
		if (list[i] == num)
			return true;
	return false;
}

void findCircle(int i)
{
    int count =0;
    for (int j = 2; j <= n; ++ j)
        if (!isIn(result,i,j) && isPrime(result[i-1] + j))
        {
            result[i] = j;
            count ++;
            findCircle(i + 1);
        }
	if (count == 0)
		return;
	if (i == n - 1 && isPrime(1 + result[n - 1]))
	{
        
		vector<int > temp ;
		for (int k = 0; k < n ; ++ k)
			temp.push_back(result[k]);
        
		results.push_back(temp);
	}
}

int main()
{
    
	int Case = 0;
	while(cin >> n)
	{
        
		Case ++;
		results.clear();
		result[0] = 1;
		findCircle(1);
        
		cout << "Case "<<Case<<":"<<endl;
		for (int i = 0; i < results.size(); ++ i )
		{
			cout << results[i][0]<< " ";

			for (int j = 1; j < results[i].size() - 1; ++ j)
				cout << results[i][j]<< " ";
            cout << results[i][results[i].size() - 1] ;//这里特别注意，最后一个不输出空格
			cout << endl;
		}        
		cout << endl;
	}
	return 0;
}