字符串相似度算法（ Levenshtein Distance算法）(zz)

昨天论坛看到的，简单写了一下
题目： 一个字符串可以通过增加一个字符，删除一个字符，替换一个字符得到另外一个字符串，假设，我们把从字符串A转换成字符串B，前面3种操作所执行的最少次数称为AB相似度
如  abc adc  度为 1
      ababababa babababab 度为 2
      abcd acdb 度为2


 字符串相似度算法可以使用 Levenshtein Distance算法(中文翻译：编辑距离算法) 这算法是由俄国科学家Levenshtein提出的。其步骤

Step	Description
1	Set n to be the length of s. Set m to be the length of t. If n = 0, return m and exit. If m = 0, return n and exit. Construct a matrix containing 0..m rows and 0..n columns.
2	Initialize the first row to 0..n. Initialize the first column to 0..m.
3	Examine each character of s (i from 1 to n).
4	Examine each character of t (j from 1 to m).
5	If s[i] equals t[j], the cost is 0. If s[i] doesn't equal t[j], the cost is 1.
6	Set cell d[i,j] of the matrix equal to the minimum of: a. The cell immediately above plus 1: d[i-1,j] + 1. b. The cell immediately to the left plus 1: d[i,j-1] + 1. c. The cell diagonally above and to the left plus the cost: d[i-1,j-1] + cost.
7	After the iteration steps (3, 4, 5, 6) are complete, the distance is found in cell d[n,m].

C++实现如下

#include <iostream>
#include <vector>
#include <string>
using namespace std;

//算法
int ldistance(const string source,const string target)
{
    //step 1

    int n=source.length();
    int m=target.length();
    if (m==0) return n;
    if (n==0) return m;
    //Construct a matrix
    typedef vector< vector<int> >  Tmatrix;
    Tmatrix matrix(n+1);
    for(int i=0; i<=n; i++)  matrix[i].resize(m+1);

    //step 2 Initialize

    for(int i=1;i<=n;i++) matrix[i][0]=i;
    for(int i=1;i<=m;i++) matrix[0][i]=i;

     //step 3
     for(int i=1;i<=n;i++)
     {
        const char si=source[i-1];
        //step 4
        for(int j=1;j<=m;j++)
        {

            const char dj=target[j-1];
            //step 5
            int cost;
            if(si==dj){
                cost=0;
            }
            else{
                cost=1;
            }
            //step 6
            const int above=matrix[i-1][j]+1;
            const int left=matrix[i][j-1]+1;
            const int diag=matrix[i-1][j-1]+cost;
            matrix[i][j]=min(above,min(left,diag));

        }
     }//step7
      return matrix[n][m];
}
int main(){
    string s;
    string d;
    cout<<"source=";
    cin>>s;
    cout<<"diag=";
    cin>>d;
    int dist=ldistance(s,d);
    cout<<"dist="<<dist<<endl;
}