hdu 2476 (string painter) 字符串刷子

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2068    Accepted Submission(s): 908

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

 

 

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题目描述 ： 给定一个初始串，目标串，每步可以通过改变一个连续的子串使其变为同一个字母，至少需要多少步？

我们发现一段序列，每一步的选择是可以改变任意长度的连续子串，

那么通过枚举改变哪些连续子串，可以包含所有的情况。

d[i]表示以i结尾的序列变成目标串需要的最少步骤。d[i]=min(d[i],d[k]+dp[k+1][i]),因为是[k+1,i]区间是连续改变的，

那么我们可以将dp[k+1][i]看成是表示[k+1,i]区间内一个相同串到目标串的最少步骤.初始化dp[i][i]=1;

dp[i][j]=dp[i][j-1]+1;

if(a[i]==a[k])   //有相同的连续改变才会有作用，不同，无论通过何种方式.每一个都需要改变，改变次数都一样

//相同的话，通过连续改变，可以减少改变次数，

dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][i-1]);

初始化d数组为0,d[i]=dp[0][1];

d[i]=min(d[i],d[k]+dp[k+1][i-1]);

通过枚举改变的连续子串的长度

动态规划： 定义状态，每一步的选择，包含了所有的可能性
最优子结构无后效性，如果状态设计不合理，会导致有后效性。

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

char a[105],b[105];

int dp[105][105],d[105];

int Length;

void init()

{

    memset(dp,0,sizeof(dp));

    memset(d,0,sizeof(d));

    for(int i=0;i<Length;i++)

        dp[i][i]=1;



}

void solve()

{

       /*   for(int i=0;i<Length;i++)

            for(int j=0;j<Length;j++)

             for(int k=i;k<=j;k++)

           {

             dp[i][j]=min(DP(dp[i][k]+dp[k+1][j]),dp[i][j]);

           }

           for(int s=0;s<Length;s++)

           {

            for(int j=0;j<Length;j++)

             printf("%d ",dp[s][j]);

               printf("\n");

           }

            printf("2\n");

        printf("%d\n",dp[0][Length-1]);

      */



       for(int t=1;t<Length;t++)

        for(int i=0;i<Length;i++)

      {

          int j=i+t;

          if(j>=Length)

            break;



          dp[i][j]=dp[i][j-1]+1;

          for(int k=i;k<j;k++)

          {

              if(b[k]==b[j])

              dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);

          }

      }



        for(int i=0;i<Length;i++)

            d[i]=dp[0][i];

        for(int i=0;i<Length;i++)

        {

            if(a[i]==b[i])

                d[i]=d[i-1];

            else

            {

                for(int k=0;k<i;k++)

                 d[i]=min(d[i],d[k]+dp[k+1][i]);

            }



        }

}

int main()

{

    //freopen("test.txt","r",stdin);

    while(~scanf("%s%s",a,b))

    {

        Length=strlen(a);

         init();

         solve();

         printf("%d\n",d[Length-1]);

    }

    return 0;

}