CodeForces - 835D Palindromic characteristics （dp）

最新推荐文章于 2018-06-04 11:19:28 发布

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最新推荐文章于 2018-06-04 11:19:28 发布

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本文解析了CodeForces D题“Palindromic Characteristics”，介绍了如何通过动态规划解决该问题，实现对字符串中各长度回文子串的计数。

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题目链接：http://codeforces.com/problemset/problem/835/D点击打开链接

Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes.

A string is 1-palindrome if and only if it reads the same backward as forward.

A string is k-palindrome (k > 1) if and only if:

Its left half equals to its right half.
Its left and right halfs are non-empty (k - 1)-palindromes.

The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string tdivided by 2, rounded down.

Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Input

The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.

Output

Print |s| integers — palindromic characteristics of string s.

        Examples
      
          input
        
abba

          output
        
6 1 0 0 

          input
        
abacaba

          output
        
12 4 1 0 0 0 0

Note

In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.

感觉描述有点不清楚啊。。

Its left half equals to its right half.这个东西我总是理解成左右得相等。。

主要的想法就是从低阶往高阶dp

dp【i】【j】为s【i~j】最高阶回文串

然后外层枚举串长度

内层从左到右

判定是否成立注意判定子条件是ij必须相差2以上

#include <bits/stdc++.h>
using namespace std;
int dp[5555][5555];
int main()
{
    string s;
    cin >> s;
    int len=s.length();

    int ans[len+1];
    for(int i=0;i<=len;i++)
        ans[i]=0;
    for(int i=0;i<len;i++)
    {
        dp[i][i]=1;
        ans[1]++;
    }
    for(int j=2;j<=len;j++)
        for(int i=0;i+j-1<=len;i++)
        {
            int r=i+j-1;
                if((dp[i+1][r-1]==0&&i+1<=r-1)||s[i]!=s[r])
                    dp[i][r]=0;
                else
                    dp[i][r]=dp[i][i+j/2-1]+1;
            if(dp[i][r])
                ans[dp[i][r]]++;
        }
    for(int i=len-1;i>=1;i--)
        ans[i]+=ans[i+1];
    for(int i=1;i<=len;i++)
        cout << ans[i] <<" ";
}