BZOJ2002 [HNOI2010]弹飞绵羊

题目蓝链

Description

\(n\)个弹簧排成一列,每一个弹簧都会给定一个向后弹射的距离,然后需要支持两个操作

  1. 查询从某个弹簧开始需要弹多少次后,就会弹到第\(n\)个弹簧之后

  2. 修改某个弹簧的向后弹射距离

Solution

考虑分块,每一个块中维护每一个位置要跳多少次才能跳出这个块,跳出去后是落在那个位置

查询直接\(\mathcal{O}(\sqrt n)\)一个块一个块的模拟去跳,修改就更新修改位置所在块的信息就可以了

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 2e5 + 10;
const int B = 400;

int n, m, blks, f[maxn], S[maxn], T[maxn];

int pos(int x) { return x <= n ? (x - 1) / B + 1 : blks + 1; }

int main() {
	freopen("sheep.in", "r", stdin);
	freopen("sheep.out", "w", stdout);

	n = read();
	for (int i = 1; i <= n; i++) f[i] = read() + i;

	blks = (n - 1) / B + 1;
	for (int i = 1; i <= blks; i++) {
		int e = min(i * B, n), tmp = (i - 1) * B;
		for (int j = e; j > tmp; j--)
			if (f[j] <= e) S[j] = S[f[j]] + 1, T[j] = T[f[j]];
			else S[j] = 1, T[j] = f[j];
	}

	m = read();
	while (m--) {
		int op = read();
		if (op == 1) {
			int x = read() + 1, ans = 0;
			while (pos(x) <= blks) ans += S[x], x = T[x];
			printf("%d\n", ans);
		}
		if (op == 2) {
			int x = read() + 1, p = pos(x);
			f[x] = read() + x;
			int tmp = (p - 1) * B, e = min(p * B, n);
			for (int i = x; i > tmp; i--) {
				if (f[i] <= e) S[i] = S[f[i]] + 1, T[i] = T[f[i]];
				else S[i] = 1, T[i] = f[i];
			}
		}
	}

	return 0;
}

Summary

这道题代码细节是真的多,我打之前没有想清一些的细节。所以我一边打一边想,调了我一个小时

一定要注意区分数组的下标和实际位置

posted @ 2018-11-18 14:04  xunzhen  阅读(53)  评论(0编辑  收藏