String painter HDU - 2476 （区间dp）

最新推荐文章于 2024-08-08 20:51:46 发布

墨碎江南

最新推荐文章于 2024-08-08 20:51:46 发布

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本文探讨了一种算法，用于解决两个等长字符串之间的转换问题，通过使用强大的字符串画家工具，将一个字符串转换为另一个字符串所需的最小操作数。输入包含多个案例，每个案例由两行组成，分别表示原始字符串A和目标字符串B。文章提供了一个C++实现示例，展示了如何计算转换所需的最少步骤。

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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

#include <stdio.h>
#include <algorithm>
#include<string.h>
using namespace std;
#define ll long long
char s1[110],s2[110];
int dp[110][110];
int ans[110];
int main()
{
    int len;
   while(~scanf("%s%s",s1,s2))
   {
       len=strlen(s1);
       memset(dp,0,sizeof(dp));
       for(int i=0;i<len;i++)  dp[i][i]=1;
       for(int k=1;k<len;k++)
       {
           for(int i=0;i+k<len;i++)
           {
               int j=i+k;
               dp[i][j]=dp[i+1][j]+1;
               for(int s=i+1;s<=j;s++)
                   if(s2[i]==s2[s])  dp[i][j]=min(dp[i][j],dp[i+1][s]+dp[s+1][j]);
           }
       }
       for(int i = 0; i<len; i++)     ans[i] = dp[0][i];
        for(int i = 0; i<len; i++)
        {
            if(s1[i] == s2[i])   ans[i] = ans[i-1];
            else
            {
                for(int j = 0; j<i; j++)
                    ans[i] = min(ans[i],ans[j]+dp[j+1][i]);
            }
        }
           printf("%d\n",ans[len-1]);

   }
}