【BZOJ5036】【JSOI2014】Palindrome

【Title link】

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【Key points of thinking】

• Insert # between every two characters (for example: " AAB "\(\Rightarrow\)" #A#A#B# ").
• Use the Manacher algorithm to find the length of the longest palindrome substring \(l_i\) centered on each character of the above string.
• Remember \(len_i=\frac{l_i+1}{2}\), it is not difficult to find that \(len_i\) is the length of the longest palindrome substring when only one side of the character is counted, that is, the number of palindrome strings .
• Now consider each query \([L,R]\), we turn it into a query \([L*2-1,R*2+1]\) in a new string, and then count only odd lengths The number of palindromic substrings of .
• Note that the answer to the original question is \(Ans\), the answer to the new question is \(New\), there is \(Ans=\frac{New-(R-L+2)}{2}\) (except \( R-L+2\) individual " # ", each string S of the original query corresponds to the two strings in the new query).
• Considering each palindrome center\(i\), if\(i-len_i+1≥L\)and\(i+len_i-1≤R\), then its contribution to the query answer is obviously\(len_i\ ), otherwise, a part of the palindrome centered on \(i\) will exceed the range of this interval, and we call \(i\) "limited".
• Remember \(Mid=\frac{L+R}{2}\), obviously the position in \([L,Mid]\) can only be "limited" by \(L\), while \([Mid+1 ,R]\) can only be "restricted" by \(R\).
• Divide the query into two parts, \([L,Mid]\) and \([Mid+1,R]\), and process them separately. Due to the symmetry of the processing method, we only discuss the \([L,Mid]\) below. part.
• What we are asking for is actually \(\sum_{i=L}^{Mid}([i-len_i+1≥L]*len_i+[i-len_i+1<L]*(i-L+1))\ ).
• Taking the query offline and sorting it according to \(L\), it is not difficult to use the tree array to obtain the value of the above formula.
• Time complexity\(O((N+Q)LogN)\).

【Code】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); }
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct querys {int pos, home; };
vector <querys> l[MAXN], r[MAXN];
char s[MAXN];
int n, q, len[MAXN];
long long ans[MAXN];
struct BinaryIndexTree {
	int n;
	long long a[MAXN];
	void init(int x) {
		n = x;
		memset(a, 0, sizeof(a));
	}
	void modify(int pos, int x) {
		for (int i = pos; i <= n; i += i & -i)
			a[i] += x;
	}
	long long query(int pos) {
		long long ans = 0;
		for (int i = pos; i >= 1; i -= i & -i)
			years += a[i];
		return ans;
	}
} Inside, Sum, Cnt;
void solvel() {
	Inside.init(n);
	Sum.init(n), Cnt.init(n);
	static vector <int> dis[MAXN];
	for (int i = 1; i <= n; i++) {
		Inside.modify(i, len[i]);
		dis[i - len[i] + 1].push_back(i);
	}
	for (int i = 1; i <= n; i++) {
		for (unsigned j = 0; j < l[i].size(); j++) {
			long long tans = 0;
			tans += Inside.query(l[i][j].pos);
			tans += Sum.query(l[i][j].pos) - Cnt.query(l[i][j].pos) * (i - 1);
			ans[l[i][j].home] += tans;
		}
		for (unsigned j = 0; j < dis[i].size(); j++) {
			int tmp = dis[i][j];
			Inside.modify(tmp, -len[tmp]);
			Sum.modify(tmp, tmp);
			Cnt.modify(tmp, 1);
		}
		Sum.modify(i, -i);
		Cnt.modify(i, -1);
	}
}
void solver() {
	Inside.init(n); int m = n + 1;
	Sum.init(n), Cnt.init(n);
	static vector <int> dis[MAXN];
	for (int i = 1; i <= n; i++) {
		Inside.modify(m - i, len[i]);
		dis[i + len[i] - 1].push_back(i);
	}
	for (int i = n; i >= 1; i--) {
		for (unsigned j = 0; j < r[i].size(); j++) {
			long long tans = 0;
			tans += Inside.query(m - r[i][j].pos);
			tans += Cnt.query(m - r[i][j].pos) * (i + 1) - Sum.query(m - r[i][j].pos);
			ans[r[i][j].home] += tans;
		}
		for (unsigned j = 0; j < dis[i].size(); j++) {
			int tmp = dis[i][j];
			Inside.modify(m - tmp, -len[tmp]);
			Sum.modify(m - tmp, tmp);
			Cnt.modify(m - tmp, 1);
		}
		Sum.modify(m - i, -i);
		Cnt.modify(m - i, -1);
	}
}
int main() {
	scanf("%s", s + 1);
	n = strlen(s + 1);
	for (int i = n; i >= 1; i--)
		s[i * 2] = s[i];
	for (int i = 0; i <= n; i++)
		s[i * 2 + 1] = '#';
	s[0] = '$', s[n * 2 + 2] = '%';
	n = n * 2 + 1;
	int pos = 0, home = 0;
	for (int i = 1; i <= n; i++) {
		if (i <= pos) {
			int j = 2 * home - i;
			len [i] = min (len [j], pos - i);
		}
		while (s[i + len[i]] == s[i - len[i]]) len[i]++;
		if (i + len[i] - 1 >= pos) {
			pos = i + len [i] - 1;
			home = i;
		}
	}
	read(q);
	for (int i = 1; i <= q; i++) {
		int x, y, z;
		read(x), read(y);
		ans[i] = -(y - x + 2);
		x = 2 * x - 1;
		y = 2 * y + 1;
		z = (x + y) / 2;
		l[x].push_back((querys) {z, i});
		r[y].push_back((querys) {z + 1, i});
	}
	solvel();
	solver();
	for (int i = 1; i <= q; i++)
		writeln(ans[i] / 2);
	return 0;
}