马拉车

先放一个链接 Link

首先，马拉车算法是为了解决最大回文串的问题而产生的，时间复杂度为 \(O(n)\)

定义一个数组 \(P_i\) , 代表以第 \(i\) 个字符为中心向外扩展所能扩展出的最大半径

定义 \(R\) 代表包含了第 \(i\) 个字符的最右边的字符串的最右边的字符的坐标

'R' means the address of the rightest letter of the axisymmetric(轴对称的) string which includes the i-th letter.

定义 \(C\) 代表包含了第 \(i\) 个字符的最右边的字符串的对称轴位置

'C' means the address of the axis of symmetry of the axisymmetric(轴对称的) string which includes the i-th letter.

It's easy to find out that R and C are in the same string.

\(Amazed!\)

那么如何求出\(P_i\)呢？

We can define \(i\_mirror\) as the symmetric letter of \(i\) of \(C\),so \(i\_mirror=2\times C - i\).

And in some cases \(P_i=P_{i\_mirror}\).

We only think about the cases in which \(P_i\) doesn't equal to \(P_{i\_mirror}\)

There are two cases.

1. 当前以 \(i\) 为中心的回文串的右端点超过了\(R\)，即\(i+P_{i\_mirror}>R\)
2. \(i\_mirror\)是因为扩展到左端点而结束的

非常优美的算法，非常简洁的代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e6 + 9;
int len,C,R,maxlen;
int P[N<<1];
char c[N],T[N<<1];
int main()
{
	scanf(" %s",c+1);
	T[0]='^';
	len=strlen(c+1);
	for(int i=1;i<=len;i++){
		T[i*2-1]='#';
		T[i*2]=c[i];
	}
	T[2*len+1]='#';
	T[2*len+2]='&';
	for(int i=2;i<=2*len;i++){
		int i_mirror;
		i_mirror=2*C-i;
		if(R>i){
			P[i]=min(R-i,P[i_mirror]);
		}
		else{
			P[i]=0;
		}
		while(T[i+P[i]+1]==T[i-P[i]-1]){
			P[i]++;
		}
		if(i+P[i]>R){
			C=i;
			R=i+P[i];
		}
	}
	for(int i=2;i<=2*len;i++){
		if(P[i]>maxlen){
			maxlen=P[i];
		}
	}
	printf("%d\n",maxlen);
	return 0;
}