HDU 2476 String painter（区间dp)

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：给你两个字符串，从a串到b串，你每次可以选择将一个区间的全部字符变成一种字符求最小到达b串的次数。

思路：学习了kuangbin大神的博客才理解的，就是先dp获得一个从空串获得b串的最小次数，这个地方，比如获得abcda的代价就是代价(bcd)+1，这里的1就是涂后面那个a的代价，因为可以一起涂成a，然后涂中间的，所以这里直接b[i] == b[k]的时候，区间dp一下。然后对a串到b串，如果a[i] == b[i]，这个i位置我我们就可以不涂，这里也区间dp一下。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int inf = 0x7fffffff, N = 1e2 + 10;
typedef long long ll;
using namespace std;
char s[N], x[N];
int dp[N][N], ans[N];
int main()
{
    while(~scanf("%s%s", s, x))
    {
        int len = strlen(s);
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i<len; i++)
            for(int j = i; j<len; j++)
                dp[i][j] = j - i + 1;
        for(int i = len - 1; i>= 0; i--)
            for(int j = i + 1; j<len; j++)
            {
                dp[i][j] = dp[i+1][j] + 1;
                for(int k = i + 1; k<=j; k++)
                    if(x[i] == x[k])
                    {
                        dp[i][j] = min(dp[i][j], dp[i+1][k-1]+dp[k][j]);
                    }
            }
        for(int i = 0; i<len; i++)
        {
            ans[i] = dp[0][i];
            if(s[i] == x[i])
            {
                if(!i) ans[i] = 0;
                else ans[i] = ans[i-1];
            }
            for(int j = 0; j<i; j++)
                ans[i] = min(ans[i], ans[j] + dp[j+1][i]);
        }
        cout<<ans[len-1]<<endl;
    }
    return 0;
}