Problem Description

There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other
character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?

Input

Input contains multiple cases. Each case consists of two lines: The first line contains string A. The second line contains string B. The length of both strings will not be greater than 100.

Output

A single line contains one integer representing the answer.

Sample Input

zzzzzfzzzzz

abcdefedcba

abababababab

cdcdcdcdcdcd

Sample Output

6

7

题意：给定两个字符串a和b，求最少须要对a进行多少次操作，才干将a变成b。每次操作时将a中随意一段变成随意一个字母所组成的段。


#include<stdio.h>

#include<string.h>

int min(int a,int b)

{

    return a>b?

b:a;

}

int main()

{

    int dp[105][105],ans[105];

    char a[105],b[105];

    while(scanf("%s%s",a,b)>0)

    {

        int len=strlen(a);

        memset(dp,0,sizeof(dp));

        for(int r=0;r<len;r++)

        for(int i=0;i<len-r;i++)

        {

            int j=i+r;

            dp[i][j]=dp[i+1][j]+1;

            for(int k=i+1;k<=j;k++)

            if(b[i]==b[k])

            dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);

        }

        for(int i=0;i<len; i++)

        ans[i]=dp[0][i];

        for(int i=0;i<len; i++)

        if(a[i]==b[i])

        {

            if(i==0)ans[i]=0;

            else ans[i]=ans[i-1];

        }

        else{

            for(int k=0;k<i;k++)

            ans[i]=min(ans[i],ans[k]+dp[k+1][i]);

        }

        printf("%d\n",ans[len-1]);

    }

}