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

D - Yet Another Palindrome Partitioning

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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.

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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.

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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.

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Sample Input 2

Copy

byebye

Sample Output 2

Copy

1

byebye can be permuted to form a palindrome byeeyb.

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Sample Input 3

Copy

abcdefghijklmnopqrstuvwxyz

Sample Output 3

Copy

26

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Sample Input 4

Copy

abcabcxabcx

Sample Output 4

Copy

3

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

#include   
using namespace std;  
string s;  
unordered_mapcnt;  
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。 
*/