String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2368 Accepted Submission(s): 1038
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
Source
2008 Asia Regional Chengdu
Recommend
lcy
对于dp[i][j] 来说最差的就是1+dp[i+1][j]咯,那怎么减少呢。。。 如果(i,j]里面也有个k使得b[i] == b[k]那么得话, 可以刷[i,k]这么长的一段,那么就有可能减少1. 但是存在问题,如果有一段跨越了k怎么办。。。 嗯,这个问题是不会出现的, 因为在[i,k]刷了b[i].如果跨越了k那么之前在b[k]刷的颜色就没了,所以是不允许有跨段的, 所以要写成&&想写成 dp[i][j] = dp[i+1][k] + dp[k+1][j]。 为什么不可以跨段 可以思考 BRBR 。 然后后来的那个 一段 要么是可以分段刷的,要么是不可以分段刷的。-
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