pat 1040. Longest Symmetric String (25)

https://www.patest.cn/contests/pat-a-practise/1040

参考了：

https://segmentfault.com/a/1190000003914228

http://blog.youkuaiyun.com/jtjy568805874/article/details/50856440

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1002;
char s[maxn];
char S[maxn * 2];
int RL[maxn * 2];

int manacher(char s[]) {
	int n = strlen(s);
	S[n + n + 2] = 0, S[0] = 2; //这里S[2n+2] 是扩充串的结尾的'\0', S[0] 是哨兵，与任何一个字符都不同，介绍后面的判断
	for (int i = 0; s[i] ; i++)
	{
		S[i + i + 1] = S[i + i + 3] = '1'; //填充字符，不与s[] 中的任何字符相同
		S[i + i + 2] = s[i];
	}
	int maxright = 0, pos = 0, maxlen = 0;
	for (int i = 1; S[i] ; i++)
	{
		/*
		RL[2*pos-i] i关于pos相对的回文半径，以i为中心的回文半径至少是这个，
		如果maxrihgt 离i足够远；否则半径只能到maxright-i
		*/
		if (i < maxright) RL[i] = min(RL[2 * pos - i], maxright - i); 
		while (S[i + RL[i]] == S[i - RL[i]]) RL[i]++;
		if (i + RL[i] > maxright) {
			maxright = i + RL[i];
			pos = i;
		}
		maxlen = max(maxlen, RL[i] - 1);
	}
	return maxlen;
}

int main()
{
	gets(s);
	printf("%d\n", manacher(s));
    return 0;
}