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.
用二进制记下前缀的每种字母奇偶性
dp[i]表示前i个最少分几段,枚举奇数字母是什么转移
可以记下每种二进制最小的dp值是什么
#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
Copy
aabxyyzz
Sample Output 1
Copy
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
Copy
byebye
Sample Output 2
Copy
1
byebye can be permuted to form a palindrome byeeyb.
Sample Input 3
Copy
abcdefghijklmnopqrstuvwxyz
Sample Output 3
Copy
26
Sample Input 4
Copy
abcabcxabcx
Sample Output 4
Copy
3
The solution is to partition s as abcabcxabcx = a + b + cabcxabcx.