hdu 2476 String painter

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4750    Accepted Submission(s): 2237

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

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

 

Sample Output

67

 

Source

2008 Asia Regional Chengdu

 

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题意是给出2个字符串，每次对1串操作可以将任意一个区间内的字符转化为相同的任意一个字母，问最少操作多少次可以将1串变成2串；
最坏情况是对1串的每一个字符进行处理，那么初始化dp数组则是按照最坏情况，然后对2串进行遍历，如果发现有相同的字符，则可以进一步优化dp数组，然后就是对1串和2串进行同时遍历，对于相同位置上的字母如果相同，则操作数是和上一长度的操作数一样的，如果不相同，则在之前长度里面找到一个位置来找最小值，这里的状态优化：ans[i]=min(ans[i],ans[j]+dp[j+1][i])。其中j为从0到i-1中找的中间位置。

#include<iostream>#include<cstring>using namespace std;int main(){    char a[111],b[111];    int dp[111][111];    int ans[111];    int len;    int i,j,k;    while(cin>>a>>b)    {        int i,j,k;        len=strlen(a);        memset(dp,0,sizeof(dp));        for(i=0;i<len;i++)        {            for(j=i;j>=0;j--)            {                dp[j][i]=dp[j+1][i]+1;                for(k=j+1;k<=i;k++)                {                    if(b[j]==b[k])                    {                      dp[j][i]=min(dp[j][i],dp[j+1][k]+dp[k+1][i]);                    }                }            }        }        for(i=0;i<len;i++)        ans[i]=dp[0][i];        for(i=0;i<len;i++)          {              if(a[i]==b[i])              ans[i]=ans[i-1];            else              {                  for(j=0;j<i;j++)                  ans[i]=min(ans[i],ans[j]+dp[j+1][i]);            }          }         cout<<ans[len-1]<<endl;    }    return 0;}