SRM500 DIV2 - SRMCards

 

Problem Statement

　　從一串數中取數，取走其中一個數的時候，要連帶取走與其相鄰的數字（例如：取走10，若數串中有9和11，就也都取走）。

　　要求：給一串數，設計算法，使得取數的次數最多。

 

Definition

Class:	SRMCards
Method:	maxTurns
Parameters:	vector <int>
Returns:	int
Method signature:	int maxTurns(vector <int> cards)
(be sure your method is public)

Constraints

-	cards will contain between 1 and 50 elements, inclusive.
-	Each element of cards will be between 1 and 499, inclusive.
-	All elements of cards will be distinct.

 

Basic Thoughts

　　爲了能讓取數的次數最多，需要如下做：

　　1. 挑出沒有相鄰項的數字，取出它；

　　2. 挑出有一個相鄰項的數字，取出它以及它的相鄰項；

　　3. 絕不允許出現取出有兩個相鄰項的數字。

 

　　所以，算法判斷時只有上述1,2兩種可能的情況。按上述兩種情況取數，直到所有數字被取完爲止。

 

class SRMCards
{
	// basic strategy
	// 1. pick up the number which has no neighbor and remove it
	// 2. pick up the number which has one neighbor and remove them two and go back to 1
public:
	int maxTurns(vector <int> cards)
	{
		int res = 0;
		int count = cards.size();
		int len = cards.size();

		int j;
		for(j=0;j<2;j=(++j)%2)
		{
			for(int i=0;i<len;i++)
			{
				if(cards[i] != -1)
				{
					vector<int>::iterator card1 = find(cards.begin(),cards.end(),cards[i]-1);
					vector<int>::iterator card2 = find(cards.begin(),cards.end(),cards[i]+1);

					if (j == 0 && card1 == cards.end() && card2 == cards.end())
					{
						res++;
						cards[i] = -1;
						count--;
					}
					else if (j == 1 && card1 == cards.end() && card2 != cards.end())
					{
						res++;
						cards[i] = -1;
						*card2 = -1;
						count-=2;
					}
					else if (j == 1 && card2 == cards.end() && card1 != cards.end())
					{
						res++;
						cards[i] = -1;
						*card1 = -1;
						count-=2;
					}
					else continue;
				}
			}

			if (count == 0)
			{
				break;
			}
		}

		return res;
	}
};