【AtCoder CODE FESTIVAL 2017 qual C】D - Yet Another Palindrome Partitioning （状压dp 字符串）

D - Yet Another Palindrome Partitioning

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, …, sN from
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

.

#include <bits/stdc++.h>
using namespace std;
string s;
unordered_map<int, int>cnt;
int main(){
cin>>s;
int cur = 0, x;
cnt[cur] = 1;
for(int i = 0; i < s.size(); ++i){
cur ^= (1 << (s[i] - 'a'));
if(cur ==  0){
continue;
}
else{
x = 1e8;
if(cnt[cur] > 0){
x= min(x, cnt[cur]);
}
for(int j = 0; j < 26; ++j){
if((cur ^ (1 << j)) == 0){ x = 1; break;}
else{
if(cnt[cur ^ (1 << j)] > 0) x = min(x, cnt[cur ^ (1 << j)] + 1);
else{
x = min(x, cnt[cur ^ (1 << (s[i] - 'a'))] + 1);
}
}
}
cnt[cur] = x;
}
}
cout << cnt[cur] << endl;
}

/*
题意：
长度2e5的字符串，问最少需要分成多少个连续的子串，在子串内部的字符可以随意排列的情况下使得这些子串都是回文串。

思路：对于子串，只要出现奇数次的字符不超过1个，该子串就是回文串。一共26个小写字母，可以用一个整数状压记录，第i
位表示该字母出现多少次。某位为0表示出现偶数次，为1表示出现奇数次。这样就很容易维护答案了。
*/