Dreamoon and Strings - CodeForces #272 (Div. 1) C dp

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本文介绍了一个字符串匹配问题，通过动态规划算法来求解最优解。具体地，算法通过寻找最佳子串来最大化目标字符串的出现次数。

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 $\text{[math]}$ 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 $\text{[math]}$ as maximum value of $\text{[math]}$ over all s' that can be obtained by removing exactly x characters froms. Dreamoon wants to know $\text{[math]}$ 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 $\text{[math]}$ 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", ""}.

题意：在第一个字符串分别删除k个字符后，第二个字符串最多出现多少次。

思路：dp[i][j]表示前i个字符删除j个后第二个字符串最多出现多少次，每次先从i向前找最近的可以形成第二个字符串的位置，此时，从j+1到i可以非连续组成第二个字符串，遍历k，dp[i][k]=dp[j][k-(i-j-len2)]+1;之后dp[i][k]=max(dp[i][k],dp[i-1][k])。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[2010][2010],ans[2010];
char s1[2010],s2[510];
int main()
{
    int i,j,k,len1,len2;
    scanf("%s%s",s1+1,s2+1);
    len1=strlen(s1+1);
    len2=strlen(s2+1);
    for(i=0;i<=len1;i++)
    {
        k=len2;
        for(j=i;j && k;j--)
           if(s1[j]==s2[k])
             k--;
        if(k==0)
          for(k=0;k<=i;k++)
             if(k-(i-j-len2)>=0 &&k-(i-j-len2)<=j)
               dp[i][k]=dp[j][k-(i-j-len2)]+1;
        for(k=0;k<=i;k++)
           dp[i][k]=max(dp[i][k],dp[i-1][k]);
    }
    for(i=0;i<=len1;i++)
       printf("%d ",dp[len1][i]);
}