2020牛客国庆集训派对day1 A.ABB

题目链接

题目描述

Fernando was hired by the University of Waterloo to finish a development project the university started some time ago. Outside the campus, the university wanted to build its representative bungalow street for important foreign visitors and collaborators.
Currently, the street is built only partially, it begins at the lake shore and continues into the forests, where it currently ends. Fernando’s task is to complete the street at its forest end by building more bungalows there. All existing bungalows stand on one side of the street and the new ones should be built on the same side. The bungalows are of various types and painted in various colors.
The whole disposition of the street looks a bit chaotic to Fernando. He is afraid that it will look even more chaotic when he adds new bungalows of his own design. To counterbalance the chaos of all bungalow shapes, he wants to add some order to the arrangement by choosing suitable colors for the new bungalows. When the project is finished, the whole sequence of bungalow colors will be symmetric, that is, the sequence of colors is the same when observed from either end of the street.
Among other questions, Fernando wonders what is the minimum number of new bungalows he needs to build and paint appropriately to complete the project while respecting his self-imposed bungalow color constraint.

输入描述:

The first line contains one integer N (1 ≤ N ≤ 4e5 ), the number of existing bungalows in the street. The next line describes the sequence of colors of the existing bungalows, from the beginning of the street at the lake. The line contains one string composed of N lowercase letters (“a” through “z”), where different letters represent different colors.

输出描述:

Output the minimum number of bungalows which must be added to the forest end of the street and painted appropriately to satisfy Fernando’s color symmetry demand.

示例1

输入

3
abb

输出

示例2

输入

12
recakjenecep

输出

示例3

输入

15
murderforajarof

输出

马拉车算法~
马拉车算法的 l[i]l[i]l[i] 记录的是以 iii 为中心的最大回文子串半径，那么这题就是当回文子串的末尾正好为字符串末尾时，去除回文子串，将字符串剩下的部分倒序添加在字符串末尾即可，答案也就是剩下部分的长度，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=4e5+5;
int n;
char p[N];
char s[N<<1];
int l[N<<1];int change(char *p){int i,len=strlen(p);s[0]='@';for(int i=1;i<=2*len;i+=2){s[i]='#';s[i+1]=p[i>>1];}s[2*len+1]='#';s[2*len+2]='$';s[2*len+3]='\0';return 2*len+1;
}int manacher(char *s,int len){int mx=0,ans=len,pos=0;for(int i=1;i<=len;i++){if(mx>i) l[i]=min(mx-i,l[2*pos-i]);else l[i]=1;while(s[i-l[i]]==s[i+l[i]]) l[i]++;if(l[i]+i>mx){mx=l[i]+i;pos=i;}if(l[i]==len-i+1) ans=min(ans,(i-l[i]+1)/2);}return ans;
}int main(){scanf("%d%s",&n,p);int len=change(p);printf("%d",manacher(s,len));return 0;
}