HDU 2476 String painter（区间DP）

String painter

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2792 Accepted Submission(s): 1272

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

这道题目完全没有思路，看了别人的题解。方法是先求空的字符数组变成目标字符数组，再求给定的字符数组变成目标字符数组。

虽然代码很短，确很值得我去思量，第二道区间DP题目

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
using namespace std;
int dp[105][105];
int ans[105];
char s1[105];
char s2[105];
int main()
{
    while(scanf("%s%s",s1+1,s2+1)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        int len=strlen(s1+1);
        for(int j=1;j<=len;j++)
        {
            for(int i=j;i>=1;i--)
            {
                dp[i][j]=dp[i+1][j]+1;
                for(int k=i+1;k<=j;k++)
                {
                    if(s2[k]==s2[i])
                        dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
                }
            }
        }
        for(int i=1;i<=len;i++)
            ans[i]=dp[1][i];
        for(int i=1;i<=len;i++)
        {
            if(s1[i]==s2[i])
                ans[i]=ans[i-1];
            else
            {
                for(int j=i-1;j>=1;j--)
                    ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
            }
        }
        printf("%d\n",ans[len]);
    }
    return 0;
}