JZOJ 2483. 【GDKOI 2021提高组DAY1】回文(palindrome)

题目

求区间最长回文串长度

\(1 \le n\le 5 \times 10^5\)

题解

比较妙的做法，主要是在询问部分

预处理出以某位为中心回文半径长 \(p_i\)，马拉车和二分+哈希均可

然后考虑询问区间 \([l..r]\)

二分一个答案半径，\(\text st\) 表维护 \([l_{new}+mid-1,r_{new}-mid+1]\) 的 \(p\) 的最大值

于是就成了判定问题

\(Code\)

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

const int N = 5e5 + 5;
int n, q, p[N << 1], lg[N << 1], f[N << 1][22];
char s[N], str[N << 1];

inline void read(int &x)
{
	x = 0; int f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
	while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	x *= f;
}

inline void manacher()
{
	str[0] = '@', str[1] = '#', n = 2;
	int len = strlen(s);
	for(register int i = 0; i < len; i++) str[n++] = s[i], str[n++] = '#';
	str[n] = '\0';
	int mx = 0, id = 0;
	for(register int i = 1; i <= n; i++)
	{
		p[i] = (i < mx ? min(p[2 * id - i], mx - i) : 1);
		while (str[i + p[i]] == str[i - p[i]]) ++p[i];
		if (i + p[i] > mx) mx = i + p[i], id = i;
	}
}

inline int query(int i, int j)
{
	if (i > j) return 0;
	int k = lg[j - i + 1];
	return max(f[i][k], f[j - (1 << k) + 1][k]);
}
inline void st_table()
{
	for(register int i = 1; i <= n; i++) f[i][0] = p[i];
	for(register int i = 2; i <= n; i++) lg[i] = lg[i - 1] + ((1 << (lg[i - 1] + 1)) == i ? 1 : 0);
	for(register int j = 1; j <= lg[n]; j++)
		for(register int i = 1; i + (1 << j) - 1 <= n; i++)
			f[i][j] = max(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
}

int main()
{
	freopen("palindrome.in", "r", stdin);
	freopen("palindrome.out", "w", stdout);
	scanf("%s%d", s, &q);
	manacher(), st_table();
	for(int l, r; q; --q)
	{
		read(l), read(r), l = (l << 1) - 1, r = (r << 1) + 1;
		int ans = 1, L = 2, R = r - l + 1, mid;
		while (L <= R)
		{
			mid = (L + R) >> 1;
			if (query(l + mid - 1, r - mid + 1) >= mid) ans = mid, L = mid + 1;
			else R = mid - 1;
		}
		printf("%d\n", ans - 1);
	}
}