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。

北大OJ_1007题：DNA Sorting_#include



作者：山丘儿