小小粉刷匠升级版(区间dp)

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): 6211    Accepted Submission(s): 2992

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 

题意：给出两个串s1和s2，一次只能将一个区间刷一次，问最少几次能让s1=s2

题解：看到区间，又是找最少操作次数，想到区间DP。

思路：直接将A转化成B不好办，我们可以先将一个空串转化成B，再比较将A直接妆化成B那个更快；

令dp[i][j]为将[i, j]区间转换成功最少操作数；

下一步的工作就是看看A直接转换成B是否更快？

令ans[i]表示A的[0, i]区间转换成B的[0, i]的最少操作；

最坏情况下，A->B,相当于空串->B;ans[i]=dp[0][i];

如果A[i]==B[i] , ans[i]=ans[i-1];这一步其实也很容易理解，i处的字符相等，那么就不需要转化了，只要转化前i-1字符；

[1, i]转换后，看看还有没有更简便的；

ans[i]=min(ans[i], ans[j]+dp[j+1][i])(1<=j<i);

#include<bits/stdc++.h>
#include<cstdio>
using namespace std;
const int MMAX=1e2+5;
char s1[MMAX],s2[MMAX];
int dp[MMAX][MMAX];
int ans[MMAX];
int main()
{
    while(~scanf("%s",s1+1))
    {
        scanf("%s",s2+1);
        int len=strlen(s1+1);
        for(int i=1;i<=len;i++) dp[i][i]=1;
        for(int le=2;le<=len;le++)  ///枚举长度
        {
            for(int l=1;l<=len-le+1;l++) ///枚举起点
            {
                int r=l+le-1; ///确定终点
                dp[l][r]=dp[l+1][r]+1; ///dp数值初始化
                for(int k=l+1;k<=r;k++)  ///枚举断点
                {
                    if(s2[l]==s2[k])
                    {
                        if(k==r) dp[l][r]=min(dp[l][r],dp[l+1][r]);
                        else dp[l][r]=min(dp[l+1][k]+dp[k+1][r],dp[l][r]);
                    }
                    else dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]);
                }
            }
        }
        ans[0]=0;
        for(int i=1;i<=len;i++)
        {
            ans[i]=dp[1][i];
            if(s1[i]==s2[i]) ans[i]=ans[i-1];
            for(int j=1;j<i;j++)
                ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
        }
        printf("%d\n",ans[len]);
    }
    return 0;
}