[HNOI2010] 弾飞绵羊

题目链接：##

传送门

题目分析：##

题外话：

我即使是死了，钉在棺材里了，也要在墓里，用这腐朽的声带喊出：

根号算法牛逼！！！

显然，这是一道LCT裸题，然而在下并不会LCT于是采用了分块瞎搞

对于每个点维护两个信息：跳出块的步数\(step[i]\)和跳出块的落点\(lo[i]\)

预处理时使用类似模拟的方法。每次只处理还未处理过的点，并且对于每次处理顺带将向后跳到过的点也处理掉，具体见代码。

对于两种操作：

• 查：利用记录的落点在大块上跳并累加答案，单次复杂度\(O(\sqrt{n})\)
• 修：分析得到，每个点维护的两个信息都只与其块内的信息有关，所以仅需要重构一遍修改点所在块即可，单次复杂度\(O(\sqrt{n})\)
  
  总复杂度约为\(O(m\sqrt{n})\)

代码：##

#include<bits/stdc++.h>
#define N 200010
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c;
	c = getchar();
	while (!isdigit(c)) {
		if (c == '-') f = -f;
		c = getchar();
	}
	while (isdigit(c)) {
		cnt = cnt * 10 + c - '0';
		c = getchar();
	}
	return cnt * f;
}
int q, pos[N], n, m, L[N], R[N], step[N], lo[N], sta[N], top, a[N], opr, x, k;
bool flag = false;

void pre_work() {
	q = sqrt(n);
	for (register int i = 1; i <= q; i++) {
		L[i] = (i - 1) * q + 1;
		R[i] = i * q;
	}
	if (R[q] < n) {
		q++;
		L[q] = R[q - 1] + 1;
		R[q] = n;
	}

	for (register int i = 1; i <= q; i++)
		for (register int j = L[i]; j <= R[i]; j++)
			pos[j] = i;

	for (register int i = 1; i <= n; i++) {
		if(step[i]) continue;
		flag = false;
		sta[++top] = i; int now = i;
		while(pos[now] == pos[now + a[now]]) {
			if (step[now + a[now]]) {
				flag = 1;
				break;
			} else {
				now += a[now];
				sta[++top] = now;
			}
		}
		int total = top + 1;
		while (top) {
			if (!flag) {
				lo[sta[top]] = now + a[now];
				step[sta[top]] = total - top;
			} else {
				lo[sta[top]] = lo[now + a[now]];
				step[sta[top]] = total - top + step[now + a[now]];
			}
			--top;
		}
	}
}
void change(int q,int k) {
	int p = pos[q];
	a[q] = k;
	for (register int i = L[p]; i <= R[p]; i++) lo[i] = step[i] = 0;
	for (register int i = L[p]; i <= R[p]; i++) {
		if(step[i]) continue;
		flag = false;
		sta[++top] = i; int now = i;
		while(pos[now] == pos[now + a[now]]) {
			if (step[now + a[now]]) {
				flag = 1;
				break;
			} else {
				now += a[now];
				sta[++top] = now;
			}
		}
		int total = top + 1;
		while (top) {
			if (!flag) {
				lo[sta[top]] = now + a[now];
				step[sta[top]] = total - top;
			} else {
				lo[sta[top]] = lo[now + a[now]];
				step[sta[top]] = total - top + step[now + a[now]];
			}
			--top;
		}
	}
}
int query(int i) {
	int v = i;
	int ans= 0;
	while (pos[v]) {
		ans += step[v];
		v = lo[v];
	}
	return ans;
}
int main() {
	n = read();
	for (register int i = 1; i <= n; i++) a[i] = read();
	m = read();
	pre_work();
	for (register int i = 1; i <= m; i++) {
		opr = read(); x = read();
		if (opr == 1) printf("%d\n", query(x + 1));
		if (opr == 2) {
			k = read();
			change(x + 1, k);
		}
	}
	return 0;
}