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  • 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[n][i]); } return 0; }


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  • 原文地址:https://www.cnblogs.com/gavanwanggw/p/6817630.html
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