标签：二进制 normal bcd stand print sel display 数字 ica
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:
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
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:
Find the minimum possible value of N when the partition satisfies the condition.
Input is given from Standard Input in the following format:
s
Print the minimum possible value of N when the partition satisfies the condition.
aabxyyzz
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.
byebye
1
byebye can be permuted to form a palindrome byeeyb.
abcdefghijklmnopqrstuvwxyz
26
abcabcxabcx
3
The solution is to partition s as abcabcxabcx = a + b + cabcxabcx.
atcoder CODE FESTIVAL 2017 qual C D - Yet Another Palindrome Partitioning
标签：二进制 normal bcd stand print sel display 数字 ica
原文地址：http://www.cnblogs.com/lengtouqing/p/7780281.html
