String painter（区间DP）

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
题目大意：
给定两个等长的字符串A，B，每次操作可将一段任意区间变为一个字符，求由A变到B的最小操作次数。

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
int dp[105][105],ans[105];///dp[i][j]中的i，j表示左右端点
int main()
{
    string a,b;
    while(cin>>a>>b)
    {
        memset(dp,0,sizeof dp);
        for(int j=0;b[j];j++)///预处理一段区间内的最小次数
            for(int i=j;i>=0;i--)
            {
                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;a[i];i++)
        {
            ans[i]=dp[0][i];
            if(a[i]!=b[i])///找更少的次数
                for(int j=0;j<i;j++)
                    ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
            else///不刷
                ans[i]=ans[i-1];
        }
        cout<<ans[b.size()-1]<<'\n';
    }
    return 0;
}