Codeforces Round #272 (Div. 1)C（字符串DP）

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

aaaaa
aa

output

2 2 1 1 0 0

input

axbaxxb
ab

output

0 1 1 2 1 1 0 0

題意：RT

思路：dp[i][j]表示s的前i個字符一共匹配了j個p串，刪掉的最少字符數

            先用一個數組en[i]預處理出在s串的每個位置i，直到能最早匹配p串的結束的位置

            轉移爲dp[ en[i+1] ][j+1]= min (dp[ en[i+1] ][j+1] ，dp[ i ][j] + (en[i+1]-i-m) )

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int MAXN = 2010;
const int INF = ~0U>>1;
char s[MAXN];
char p[510];
int en[MAXN];
int dp[MAXN][MAXN];

int main()
{
    scanf("%s%s",s,p);
    int n=strlen(s);
    int m=strlen(p);

    for(int i=0;i