HDU String painter

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?

InputInput 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.
OutputA single line contains one integer representing the answer.Sample Input

zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd

Sample Output

6
7



思路:我们考虑到将空白串刷成2串的代价 = 将当前第i号所在的区间i,j 刷成和 2 串相同的代价(一串的i！=2串的i)
f[i][j]表示将空白串刷成2串的代价 考虑 i状态
如果枚举断点k 串2 的 i == 串2 的 k 那么就可以不用更新

同理 
ans[i] 表示 1到i 区间 所耗费最小代价

如果 i 号的串相等 那么就不用刷
如果不相等 那么就枚举断点 后面是刷空串的代价

code:

//
#include<bits/stdc++.h>
using namespace std;
#define maxnn 200
int f[maxnn][maxnn],ans[maxnn];
int n;
string str1,str2;
int main()
{
      while(cin>>str1>>str2)
    {
        int n=str2.size();
        
        for(int i=0;i<n;i++)
        f[i][i]=1;
        for(int i=n-2;i>=0;i--)
        for(int j=i+1;j<n;j++)
        {
            f[i][j]=f[i+1][j]+1;
            for(int k=i+1;k<=j;k++)
                if(str2[k]==str2[i])
                    f[i][j]=min(f[i][j],f[i+1][k-1]+f[k][j]);
        }
        for(int i=0;i<=n;i++)
        {
            ans[i]=f[0][i];
            if(str1[i]==str2[i])
            {
                if(i==0)ans[i]=0;
                else
                ans[i]=ans[i-1];
            }
            for(int k=0;k<i;k++)
            {
                ans[i]=min(ans[i],ans[k]+f[k+1][i]);
            }
        }
        cout<<ans[n-1]<<endl;
    }
}