昨天論壇看到的，簡單寫了一下
題目： 一個字符串可以通過增加一個字符，刪除一個字符，替換一個字符得到另外一個字符串，假設，我們把從字符串A轉換成字符串B，前面3種操作所執行的最少次數稱為AB相似度
如 abc adc 度為 1
ababababa babababab 度為 2
abcd acdb 度為2
字符串相似度算法可以使用 Levenshtein Distance算法(中文翻譯：編輯距離算法) 這算法是由俄國科學家Levenshtein提出的。其步驟
| 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++實現如下
#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;
}
#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;
}