学步园

AGTC

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

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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

A 
G 
T 
A 
A 
G 
T 
A 
G 
G 
C

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| 
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A 
G 
T 
C 
T 
G 
* 
A 
C 
G 
C

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

4

设有长度
l1

和
l2

的字符串
s1

和
s2,res[i][j]

为
s1

的前
i

个字符变成
s2

的前
j

个字符所需的最少操作数
,

转移方程为
:

if (s1[i]==s2[j]) res[i+1][j+1]=min(res[i][j],res[i+1][j]+1,res[i][j+1]+1);

else res[i+1][j+1]=min(res[i][j]+1,res[i+1][j]+1,res[i][j+1]+1);

当
(s1[i]==s2[j])

时
,res[i+1][j+1]

可以直接取
res[i][j],

也可以从
res[i+1][j]

加
1

得到
,

也可以从
res[i][j+1]

得到
,

取其中的最小者便是当前的最优解
.

(s1[i]!=s2[j])

时相似
.

代码如下
:

#include<stdio.h><br /> #include<string.h><br /> #define MAXN 1005<br /> int min(int a,int b,int c)<br /> {<br /> if (a>b) return b>c?c:b;<br /> return a>c?c:a;<br /> }<br /> int res[MAXN][MAXN];<br /> int main()<br /> {<br /> int l1,l2,i,j;<br /> char s1[MAXN],s2[MAXN];<br /> while (scanf("%d%s",&l1,s1)!=EOF)<br /> {<br /> scanf("%d%s",&l2,s2);<br /> for (i=0;i<=l1;++i) res[i][0]=i;<br /> for (i=0;i<=l2;++i) res[0][i]=i;<br /> for (i=0;i<l1;++i)<br /> for (j=0;j<l2;++j)<br /> if (s1[i]==s2[j]) res[i+1][j+1]=min(res[i][j],res[i+1][j]+1,res[i][j+1]+1);<br /> else res[i+1][j+1]=min(res[i][j]+1,res[i+1][j]+1,res[i][j+1]+1);<br /> printf("%d/n",res[l1][l2]);<br /> }<br /> return 0;<br /> }<br />