Problem Statement

We have a string s consisting of lowercase English letters. Snuke is partitioning s into some number of non-empty substrings. Let the subtrings obtained be s1, s2, …, sNfrom left to right. (Here, s=s1+s2+…+sN holds.) Snuke wants to satisfy the following condition:

• For each i
  (1≤i≤N), it is possible to permute the characters in si and obtain a palindrome.

Find the minimum possible value of N when the partition satisfies the condition.

用二進制記下前綴的每種字母奇偶性

dp[i]表示前i個最少分幾段,枚舉奇數字母是什麽轉移

可以記下每種二進制最小的dp值是什麽

#include<bits/stdc++.h>  
using namespace std;
 
char s[200005];
int f[1<<26];
int main(){
    scanf("%s",&s);
    int n=strlen(s),td;
    for (int i=1;i<(1<<26);i++) f[i]=1000000000;
    for (int i=0,zt=0;i<n;i++){
        zt^=(1<<(s[i]-‘a‘));
        td=f[zt]+1;
        for (int j=0;j<26;j++) td=min(td,f[zt^(1<<j)]+1
 
);
        f[zt]=min(f[zt],td);
    }
    printf("%d",td);
}

Time limit : 3sec / Memory limit : 512MB

Score : 700 points

Problem Statement

We have a string s consisting of lowercase English letters. Snuke is partitioning s into some number of non-empty substrings. Let the subtrings obtained be s1, s2, …, sNfrom left to right. (Here, s=s1+s2+…+sN holds.) Snuke wants to satisfy the following condition:

• For each i (1≤i≤N), it is possible to permute the characters in si and obtain a palindrome.

Find the minimum possible value of N when the partition satisfies the condition.

Constraints

• 1≤|s|≤2×105
• s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum possible value of N when the partition satisfies the condition.

Sample Input 1

aabxyyzz

Sample Output 1

2

The solution is to partition s as aabxyyzz = aab + xyyzz. Here, aab can be permuted to form a palindrome aba, and xyyzz can be permuted to form a palindrome zyxyz.

Sample Input 2

byebye

Sample Output 2

1

byebye can be permuted to form a palindrome byeeyb.

Sample Input 3

abcdefghijklmnopqrstuvwxyz

Sample Output 3

26

Sample Input 4

abcabcxabcx

Sample Output 4

3

The solution is to partition s as abcabcxabcx = a + b + cabcxabcx.

atcoder CODE FESTIVAL 2017 qual C D - Yet Another Palindrome Partitioning