LOJ6279 数列分块入门3

LOJ6279 数列分块入门 3

标签

• 分块入门

前言

• 犯了一些低级错误，debug了半天~

简明题意

• 维护序列，支持两种操作：
  1. 区间加
  2. 查询某个数的前驱

思路

• 这题和数列分块入门 2基本差不多了，就是查询的时候稍微改变一下就可以了QAQ
• 具体就是整块的直接二分，不整块的暴力找，然后取最大的就行了

注意事项

• 一定要注意细节。tag不要忘记加了

总结

• 分块好玩又简单QAQ，为啥没有早点学呢

AC代码

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;

const int maxn = 1e5 + 10;

int n, a[maxn];
int pos[maxn], len, tag[maxn];
vector<int> b[maxn];

void reset(int id)
{
	b[id].clear();
	for (int i = (id - 1) * len + 1; i <= min(id * len, n); i++)
		b[id].push_back(a[i]);
	sort(b[id].begin(), b[id].end());
}

void change(int l, int r, int c)
{
	for (int i = l; i <= min(pos[l] * len, r); i++)
		a[i] += c;
	reset(pos[l]);

	if (pos[l] != pos[r])
	{
		for (int i = r; i >= (pos[r] - 1) * len + 1; i--)
			a[i] += c;
		reset(pos[r]);
	}

	for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
		tag[i] += c;
}

int cal(int l, int r, int c)
{
	int ans = -1;

	for (int i = l; i <= min(pos[l] * len, r); i++)
		if (a[i] + tag[pos[i]] < c)
			ans = max(ans, a[i] + tag[pos[i]]);

	if (pos[l] != pos[r])
		for (int i = r; i >= max((pos[r] - 1) * len + 1, l); i--)
			if (a[i] + tag[pos[i]] < c)
				ans = max(ans, a[i] + tag[pos[i]]);

	for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
	{
		auto t = lower_bound(b[i].begin(), b[i].end(), c - tag[i]);
		if (t != b[i].begin() && *(t - 1) + tag[i] < c)
			ans = max(ans, *(t - 1) + tag[i]);
	}

	return ans;
}

void solve()
{
	scanf("%d", &n);
	len = sqrt(n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]), b[pos[i] = (i - 1) / len + 1].push_back(a[i]);

	//预处理
	for (int i = 1; i <= pos[n]; i++)
		sort(b[i].begin(), b[i].end());

	for (int i = 1; i <= n; i++)
	{
		int opt, l, r, c;
		scanf("%d%d%d%d", &opt, &l, &r, &c);
		if (opt == 0)
			change(l, r, c);
		else
			printf("%d\n", cal(l, r, c));
	}
}

int main()
{
	freopen("Testin.txt", "r", stdin);
	freopen("Testout.txt", "w", stdout);
	solve();
	return 0;
}