-1

以下是我编写 BK-Tree 的尝试，对于150000它需要的 word 文件8 seconds

有什么办法可以减少这个时间。

以下是我的代码

#include <stdio.h>
#include <string>
#include <vector>
#include <fstream>  
#include <iostream>  
#include <sstream>


#include "Timer.h"

class BkTree {
public:
    BkTree();
    ~BkTree();
    void insert(std::string m_item);
private:
    size_t EditDistance( const std::string &s, const std::string &t );
    struct Node {
        std::string m_item;
        size_t m_distToParent;
        Node *m_firstChild;
        Node *m_nextSibling;
        Node(std::string x, size_t dist);       
        ~Node();
  };
  Node *m_root;
  int   m_size;
protected:
};

BkTree::BkTree() {
    m_root = NULL; 
    m_size = 0;
}

BkTree::~BkTree() { 
    if( m_root ) 
        delete m_root; 
}

BkTree::Node::Node(std::string x, size_t dist) {
    m_item         = x;
    m_distToParent = dist;
    m_firstChild   = m_nextSibling = NULL;
}

BkTree::Node::~Node() {
    if( m_firstChild ) 
        delete m_firstChild;
    if( m_nextSibling ) 
        delete m_nextSibling;
}

void BkTree::insert(std::string m_item) {
    if( !m_root ){
        m_size = 1;
        m_root = new Node(m_item, -1);
        return;
    }
    Node *t = m_root;
    while( true ) {
        size_t d = EditDistance( t->m_item, m_item );
        if( !d ) 
            return;
        Node *ch = t->m_firstChild;
        while( ch ) {
            if( ch->m_distToParent == d ) { 
                t = ch; 
                break; 
            }
            ch = ch->m_nextSibling;
        }
        if( !ch ) {
            Node *newChild = new Node(m_item, d);
            newChild->m_nextSibling = t->m_firstChild;
            t->m_firstChild = newChild;
            m_size++;
            break;
        }
    }
}

size_t BkTree::EditDistance( const std::string &left, const std::string &right ) {
    size_t asize = left.size();
    size_t bsize = right.size();
    std::vector<size_t> prevrow(bsize+1);
    std::vector<size_t> thisrow(bsize+1);

    for(size_t i = 0; i <= bsize; i++)
        prevrow[i] = i;

    for(size_t i = 1; i <= asize; i ++) {
        thisrow[0] = i;
        for(size_t j = 1; j <= bsize; j++) {
            thisrow[j] = std::min(prevrow[j-1] + size_t(left[i-1] != right[j-1]), 
                                  1 + std::min(prevrow[j],thisrow[j-1]) );
        }
        std::swap(thisrow,prevrow);
    }
    return prevrow[bsize];
}


void trim(std::string& input_str) {
      if(input_str.empty()) return;
      size_t startIndex = input_str.find_first_not_of(" ");
      size_t endIndex = input_str.find_last_not_of("\r\n");
      std::string temp_str = input_str;
      input_str.erase();
      input_str = temp_str.substr(startIndex, (endIndex-startIndex+ 1) );
}


int main( int argc, char **argv ) {
    BkTree *pDictionary = new BkTree();

    std::ifstream dictFile("D:\\dictionary.txt");

    Timer *t = new Timer("Time Taken to prepare Tree = ");
    std::string line;   
    if (dictFile.is_open()) {
        while (! dictFile.eof() ) {
            std::getline (dictFile,line);
            trim(line);
            pDictionary->insert(line);
        }
        dictFile.close();
    }
    delete t;
    delete pDictionary;
    return 0;
}




class Timer {
public:
    Timer (const std::string &name = "undef");
    ~Timer (void);
private:
    std::string m_name;
    std::clock_t m_started;
protected:
};


Timer::Timer (const std::string &name) : m_name(name), m_started(clock()) {
}
Timer::~Timer (void) {
    double secs = static_cast<double>(std::clock() - m_started) / CLOCKS_PER_SEC;
    std::cout << m_name << ": " << secs << " secs." << std::endl;
}
4

1 回答 1

1

您可以通过消除 I/O 来缩短时间。要测试您的算法，请从等式中删除不受程序直接控制的尽可能多的对象。例如，操作系统控制 I/O，它不在您的控制范围内。常量文本数组消除了很多操作系统的参与（操作系统仍然可以根据操作系统内存分配对数组进行 分页）。

其次，大多数树结构都是面向数据的。它们的性能时间取决于数据。尝试三组数据：升序排序、“随机”和降序排序。注意每个的时间。

查看您的循环并分解出任何常量。在循环中创建临时变量，以便在内部循环中进行常量计算。删除不必要的操作。

最后，如果您的程序和算法非常健壮，请在其他项目上工作。仅在必要时进行优化。

于 2011-05-26T17:09:20.000 回答