poj 3356

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7934		Accepted: 3147

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

A  G  T  A  A  G  T  A  G  G  C
|  |  |        |     |     |  |
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

//t[i][j]表示第一个字符串的前i个字符要和第二个字符串的前j个字符匹配需要的最少操作次数   
//初始化t[i][0]=i,若第二个字符串为0，只好把第一个字符串全删掉，所以t[i][0]=i   
//同理，t[0][j]=j   
//状态转移方程：若s1[i-1]==s2[j-1],则不需要操作，那么t[i][j]=t[i-1][j-1]   
//否则，我们可能有三步操作，删除，插入，变换，所以t[i][j]=min(t[i-1][j]+1,t[i][j-1]+1,t[i-1][j-1]+1)   
//把删除和插入看作是一个操作，自然t[i-1][j-1]+1对应的是变换了。
#include<stdio.h>
#include<string.h>

char a[1010];
char b[1010];
int t[1010][1010],A,B;

int main()
{
    int i,j;
    while(scanf("%d%s%d%s",&A,a,&B,b)!=EOF)
    {
        memset(t,0,sizeof(t));
        for(i=0;i<=A;i++)
            t[i][0]=i;
        for(j=0;j<=B;j++)
            t[0][j]=j;

        for(i=1;i<=A;i++)
            for(j=1;j<=B;j++)
            {
                if(a[i-1]==b[j-1]) t[i][j]=t[i-1][j-1];
                else 
                {
                    t[i][j]=t[i-1][j]+1<t[i-1][j-1]+1?t[i-1][j]+1:t[i-1][j-1]+1;
                    t[i][j]=t[i][j]<t[i][j-1]+1?t[i][j]:t[i][j-1]+1;
                }
            }
            printf("%d\n",t[A][B]);
    }
    return 0;
}