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Codeforces Round #272 (Div. 1)C(字符串DP)_html/css_WEB-ITnose

C. Dreamoon and Strings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output dreamoon has a string s and a pattern string p. he first removes exactly x characters from s obtaining string s' as a result. then he calculates  that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. he wants to make this number as big as possible. More formally, let's define  as maximum value of  over all s' that can be obtained by removing exactly x characters froms. Dreamoon wants to know  for all x from 0 to |s| where |s| denotes the length of string s. Input The first line of the input contains the string s (1?≤?|s|?≤?2?000). The second line of the input contains the string p (1?≤?|p|?≤?500). Both strings will only consist of lower case English letters. 立即学习 “ 前端免费学习笔记(深入) ”; Output Print |s|?+?1 space-separated integers in a single line representing the  for all x from 0 to |s|. Sample test(s) input
aaaaaaa
output
2 2 1 1 0 0
input
axbaxxbab
output
0 1 1 2 1 1 0 0
题意:RT 思路:dp[i][j]表示s的前i个字符一共匹配了j个p串,删掉的最少字符数 先用一个数组en[i]预处理出在s串的每个位置i,直到能最早匹配p串的结束的位置 转移为dp[ en[i+1] ][j+1]= min (dp[ en[i+1] ][j+1] ,dp[ i ][j] + (en[i+1]-i-m) )

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