E. Dreamoon and Strings（Codeforces Round #272)

E. 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

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters
from s are {"aaaaa", "aaaa","aaa", "aa", "a", ""}.

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab","a", ""}.

dp[i][j] 为在字符串s的前i个删j个字符。k为从i開始，删除k个字符，会多出来一个字符串p。

则dp[i][j]=max(dp[i][j],dp[i-m-k][j-k]+1),m为字符串p的长度。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
const int maxn=2000+100;
using namespace std;
char s1[maxn],s2[maxn];
int dp[maxn][maxn];
int n,m;
int solve(int i)
{
int top=m,ans=0;
if(i<m)
return maxn;
while(top&&i)
{
if(s1[i]==s2[top])
top--;
else
ans++;
i--;
}
if(top)
return maxn;
else
return ans;
}
int main()
{
scanf("%s%s",s1+1,s2+1);
n=strlen(s1+1);
m=strlen(s2+1);
for(int i=0;i<=n;i++)
for(int j=i+1;j<=n;j++)
dp[i][j]=-maxn;//j>i不可能有值。赋以无穷小，防止被取到
for(int i=1;i<=n;i++)
{
int k=solve(i);
for(int j=0;j<=i;j++)
{
dp[i][j]=max(dp[i-1][j],dp[i][j]);
if(j>=k)
{
dp[i][j]=max(dp[i][j],dp[i-m-k][j-k]+1);//前面的j>i赋无穷小就是防止j-k>i-m-k时被取到。

}
}
}
for(int i=0;i<=n;i++)
{
printf("%d ",dp
[i]);
}
return 0;
}