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  • Codeforces Round #272 (Div. 1) Problem C. Dreamoon and Strings

    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
    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", ""}.

    题解报告:

    几天后补上

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    #include <vector>
    #include <stack>
    #include <map>
    #include <set>
    #include <queue>
    #include <iomanip>
    #include <string>
    #include <ctime>
    #include <list>
    #include <bitset>
    typedef unsigned char byte;
    #define pb push_back
    #define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
    #define local freopen("in.txt","r",stdin)
    #define pi acos(-1)
    
    using namespace std;
    const int maxn = 2e3 + 50;
    char s[maxn],p[maxn];
    int errorcode,ans[maxn];
    
    inline void updata(int & x ,int v){x=min(x,v);}
    
    int main(int argc,char *argv[])
    {
        scanf("%s%s",s+1,p+1);
        int l1 = strlen(s+1),l2=strlen(p+1);
        int dp[l1+3][l2+3][l1/l2+2],shang=l1/l2+1;
        memset(dp,0x3f,sizeof(dp));errorcode=dp[0][0][0];dp[0][0][0]=0;memset(ans,0,sizeof(ans));
        for(int i = 0 ; i < l1 ; ++ i)
            for(int j = 0 ; j < l2 ; ++ j)
                for(int k = 0 ; k <= shang ; ++ k)
                    if(dp[i][j][k]!=errorcode)
                    {
                        if(j==0) updata(dp[i+1][j][k],dp[i][j][k]);
                        else updata(dp[i+1][j][k],dp[i][j][k]+1);
                        if(s[i+1]==p[j+1])
                        {
                            if(j==l2-1)
                                updata(dp[i+1][0][k+1],dp[i][j][k]);
                            else
                                updata(dp[i+1][j+1][k],dp[i][j][k]);
                        }
                        else
                        {
                            updata(dp[i+1][0][k],dp[i][j][k]);
                        }
                    }
        for(int j = 0 ; j <= l2 ; ++ j)
            for(int k = 0 ; k <= shang ; ++ k)
                if(dp[l1][j][k] != errorcode)
                 {
                    int v = dp[l1][j][k];
                    ans[v] = max(ans[v],k);
                 }
        for(int i = 0 ; i <= l1 ; ++ i)
            if(ans[i])
            {
                for(int j = i + 1 ; j <= l1 ; ++ j)
                    ans[j]=max(ans[j],min((l1-j)/l2,ans[i]));
            }
        printf("%d",ans[0]);
        for(int i = 1 ; i <= l1 ; ++ i) printf(" %d",ans[i]);printf("
    ");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Xiper/p/4884008.html
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