AGTC(动态规划-最短编辑距离)

╰半橙微兮° 2022-08-02 14:49 179阅读 0赞

AGTC

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;10742</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4133</td> 
  </tr> 
 </tbody> 
</table>

Description

Let *x* and *y* be two strings over some finite alphabet *A*. We would like to transform *x* into *y* allowing only operations given below:

*  **Deletion:** a letter in *x* is missing in *y* at a corresponding position.
 *  **Insertion:** a letter in *y* is missing in *x* at a corresponding position.
 *  **Change:** letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

> **Illustration**
> 
>     A G T A A G T * A G G C
>     
>     | | |       |   |   | |
>     
>     A G T * C * T G A C G C
> 
> **Deletion:** \* in the bottom line  
> **Insertion:** \* in the top line  
> **Change:** when the letters at the top and bottom are distinct

This tells us that to transform *x* = AGTCTGACGC into *y* = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings *x* and *y* to be fixed, such that the number of letters in *x* is *m* and the number of letters in *y* is *n* where *n* ≥ *m*.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string *x* into a string *y*.

Input

The input consists of the strings *x* and *y* prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string *x* into a string *y*.

Sample Input

10 AGTCTGACGC
    11 AGTAAGTAGGC

Sample Output

Source

[Manila 2006][]

#include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define N 1010
    int dp[N][N];//dp[i][j]表示a的前i个和b的前j个相同后的最短距离
    char a[N],b[N];
    void match(char *s,char *s1,int lena,int lenb)
    {
        int i,j,inserts,deletes,replaces;
        for(i=1;i<=lena;i++)
        {
            dp[i][0]=i;
        }
        for(j=1;j<=lenb;j++)
        {
            dp[0][j]=j;
        }
        for(i=1;i<=lena;i++)
        {
            for(j=1;j<=lenb;j++)
            {
                inserts=dp[i][j-1]+1;//插入
                deletes=dp[i-1][j]+1;//删除
                replaces;//替换
                if(s[i-1]==s1[j-1])
                {
                    replaces=dp[i-1][j-1];
                }
                else
                {
                    replaces=dp[i-1][j-1]+1;
                }
                dp[i][j]=min(min(inserts,deletes),replaces);
            }
        }
        cout<<dp[lena][lenb]<<endl;
    }
    int main()
    {
        int la,lb;
        while(cin>>la>>a)
        {
            cin>>lb>>b;
            match(a,b,la,lb);
        }
        return 0;
    }

[Manila 2006]: http://poj.org/searchproblem?field=source&key=Manila+2006