北大OJ_1007题：DNA Sorting

该题的意思是输入指定数量的字符串，每个字符串的长度一样，找出每个字符串中逆序对，然后按逆序对的升序输出所以的字符串，逆序对相同的则按输入时的顺序输出。

此题的突破点在找逆序对，以下列举两种找出逆序对的方法。

穷举法找逆序对（时间复杂度为O(n^2)）

#include <iostream>
#include <string>
#include <deque>
#include <cmath>

using namespace std;

struct tagString
{
	int nInvertedCount;
	string strSequence;
};

int main()
{
	deque<tagString> mStrList;
	tagString mTagString;
	short n,m;
	cin >> n >> m;
	while( m > 0 )
	{
		cin >> mTagString.strSequence;
		int nInvertedCount = 0;
		for( string::size_type i = 0; i < mTagString.strSequence.size() - 1; ++i )
		{
			for( string::size_type j = i + 1; j < mTagString.strSequence.size(); ++j )
			{
				if ( mTagString.strSequence[i] > mTagString.strSequence[j] )
				{
					nInvertedCount++;
				}
			}
		}

		mTagString.nInvertedCount = nInvertedCount;
		if ( mStrList.empty() )
		{
			mStrList.push_back( mTagString );
		}
		else
		{
			int i = mStrList.size()-1;
			for ( ; i >= 0; --i )
			{
				if ( mStrList[i].nInvertedCount > nInvertedCount )
				{
					continue;
				}
				break;
			}
			mStrList.insert( mStrList.begin() + (i + 1), mTagString );
		}

		--m;
	}

	for ( deque<tagString>::size_type i = 0; i < mStrList.size(); ++i )
	{
		cout << mStrList[i].strSequence << endl;
	}

	return 0;
}

归并排序找逆序对（时间复杂度为O(nlogn)）

#include <iostream>
#include <string>
#include <deque>
#include <cmath>

using namespace std;

int Merge( string& strSrc, int low, int mid, int high )
{
	int i = low;
	int j = mid + 1;
	int nCount = 0;
	string strTemp( high - low + 1, 0 );
	int k = 0;

	while( i <= mid && j <= high )
	{
		if ( strSrc[i] <= strSrc[j] )
		{
			strTemp[k++] = strSrc[i++];
		}
		else
		{
			strTemp[k++] = strSrc[j++];
			nCount += j - k - low;
		}
	}

	while( i <= mid )
	{
		strTemp[k++] = strSrc[i++];
	}
	while( j <= high )
	{
		strTemp[k++] = strSrc[j++];
	}

	for ( k = 0; k < (high-low+1); ++k )
	{
		strSrc[low+k] = strTemp[k];
	}

	return nCount;
}

int MergeSort( string& strSrc, int low, int high )
{
	int nCount = 0;
	if ( low < high )
	{
		int mid = ( low + high ) / 2;
		nCount += MergeSort( strSrc, low, mid );
		nCount += MergeSort( strSrc, mid + 1, high );
		nCount += Merge( strSrc, low, mid, high );
	}

	return nCount;
}

struct tagString
{
	int nInvertedCount;
	string strSequence;
};

int main()
{
	deque<tagString> mStrList;
	tagString mTagString;
	short n,m;
	cin >> n >> m;
	while( m > 0 )
	{
		cin >> mTagString.strSequence;
		string strTemp = mTagString.strSequence;
		mTagString.nInvertedCount = MergeSort( strTemp, 0, strTemp.size()-1 );
		if ( mStrList.empty() )
		{
			mStrList.push_back( mTagString );
		}
		else
		{
			int i = mStrList.size()-1;
			for ( ; i >= 0; --i )
			{
				if ( mStrList[i].nInvertedCount > mTagString.nInvertedCount )
				{
					continue;
				}
				break;
			}
			mStrList.insert( mStrList.begin() + (i + 1), mTagString );
		}

		--m;
	}

	for ( deque<tagString>::size_type i = 0; i < mStrList.size(); ++i )
	{
		cout << mStrList[i].strSequence << endl;
	}

	return 0;
}

总结

理论上来说，第二种方法要快一些，但是第二种方法里在排序的时候存在很多的交换，在OJ上编译运行反而是第一种方法更快。

作者：山丘儿

转载请标明出处，谢谢。原文地址：http://blog.csdn.net/s634772208/article/details/46568015