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