String painter(区间dp)

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

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题意：

给出两个字符串，现，可以对第一个字符串进行修改操作——将某段连续区间全部修改为同一个字符。问由第一个字符串$A$变成第二个串$B$，至少需要做几次操作？

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分析：

由$A$变成$B$，对于当前第$i$位，有以下转移：

• 如果 $A[i]==B[i]$，那么第 $i$ 位可以不修改，即 $$f[i] = f[i-1]$$
  • 否则，在$[1,k]$的基础上，修改$[k+1,i]$，即 $$f[i] = f[k]+change(k+1,i)$$

  那么，现在的问题就是求解 $change(k+1,i)$，由于对于任意一个 $k$ 都需要求解，问题就相当于，将一个空串变成第二个字符串的最少操作次数

  我们可以定义状态： $dp[i][j]$ 为 修改区间$[i,j]$变成$B_i,B_{i+1},...,B_j$需要的最少操作次数

  对于区间 $[i,j]$ , $dp[i][j]$转移如下：

  • 在区间 $[i+1,j]$ 的基础上，仅仅修改 $i$ 位置的字符，则有 $$dp[i][j] = min( dp[i][j] , dp[i+1][j] + 1-(B[i]==B[j]))$$
  • 分成两段区间 $[i,k]$ 和 $[k+1,j]$ 进行修改，则有 $$dp[i][j] = min(dp[i][j] , dp[i][k]+dp[k+1][j])$$

  则可以将转移方程$f[i] = f[k]+change(k+1,i)$ 修改为 $f[i] = f[k]+dp[k+1][i]$

  最终答案为 $f[n]$

  ---

  代码：

  #include<stdio.h>
  #include<string.h>
  #include<stdlib.h>
  #include<math.h>
  #include<algorithm>
  #include<time.h>
  #include<iostream>
  #include<vector>
  #include<stack>
  #include<queue>
  #include<set>
  #include<map>
  
  using namespace std;
  
  #define bll long long
  
  const int maxn = 110;
  char s[maxn],t[maxn];
  int dp[maxn][maxn],ans[maxn];
  
  int main()
  {
      while (scanf("%s %s",s+1,t+1)!=EOF)
      {
          int n = strlen(s+1);
          memset(dp,0,sizeof(dp));
          for (int i=1;i<=n;i++)
             for (int j=i;j<=n;j++) dp[i][j] = j-i+1;
  
          for (int i=n;i>0;i--)
          {
              for (int j=i+1;j<=n;j++)
              {
                  int tmp = (t[i]==t[j])? 0 : 1;
                  dp[i][j] = min(dp[i][j],dp[i+1][j]+tmp);
                  for (int k=i+1;k<j;k++)
                      dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]);
              }
          } 
  
          memset(ans,0,sizeof(ans));
          for (int i=1;i<=n;i++)
          {
              ans[i] = (s[i]==t[i])? ans[i-1] : ans[i-1]+1;
              for (int j=0;j<i;j++)
                  ans[i] = min(ans[i],ans[j]+dp[j+1][i]);
          }
          printf("%d\n",ans[n]);
      }
      return 0;
  }