String painter（DP综合题：区间DP（两次DP））-优快云博客

本文介绍了一道关于字符串转换的问题，旨在寻找将一个字符串通过最少的操作次数转换为另一个字符串的方法。文章提供了完整的AC代码实现，使用了动态规划来解决这个问题。

Link：http://acm.hdu.edu.cn/showproblem.php?pid=2476

String painter

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2543 Accepted Submission(s): 1145

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

Source

2008 Asia Regional Chengdu

AC code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<cmath>
#include<vector>
#include<string.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
int dp2[111][111];
int dp[111];
char s1[111],s2[111];
int main()
{
    //freopen("D:\\in.txt","r",stdin);
    int T,cas,i,j,k,n,m,len;
    while(scanf("%s%s",s1+1,s2+1)!=EOF)
    {
       len=strlen(s2+1);
       memset(dp2,0,sizeof(dp2));
       memset(dp,0,sizeof(dp));
       for(i=1;i<=len;i++)
        dp2[i][i]=1;
       for(i=len-1;i>=1;i--)
       {
           for(j=i+1;j<=len;j++)
           {
                //dp2[i][j]=min(dp2[i+1][j]+(s2[i]==s2[i+1]?0:1),dp2[i+1][j-1]+(s2[j]==s2[j-1]?0:1));
                dp2[i][j]=dp2[i+1][j]+1;
                for(k=i+1;k<=j;k++)
                {
                    if(s2[i]==s2[k])
                    dp2[i][j]=min(dp2[i][j],dp2[i+1][k-1]+dp2[k][j]);
                }
           }
       }
       for(i=1;i<=len;i++)
       {
           dp[i]=dp2[1][i];
           if(s1[i]==s2[i])
           {
               if(i==1)
                dp[i]=0;
               else
                dp[i]=dp[i-1];
           }
           else
           {
                for(j=1;j<i;j++)
                {
                    dp[i]=min(dp[i],dp[j]+dp2[j+1][i]);
                }
           }
       }
       printf("%d\n",dp[len]);
    }
    return 0;
}