bzoj1935: [Shoi2007]Tree 园丁的烦恼lowbit 离散化

链接

bzoj

最好不要去luogu，数据太水

思路

一个询问转化成四个矩阵，求起点\((0,0)到(x,y)\)的矩阵

离线处理，离散化掉y，x不用离散。

一行一行的求，每次处理完一行之后下一行的贡献直接叠加到当前。

用lowbit统计

错误

离散化小心点，是y-1不是y

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7, maxn = 5e5 + 1;
int read() {
	int x = 0, f = 1; char s = getchar();
	for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
	for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
	return x * f;
}

int n, m, ans[N], x[N], y[N], X[N], Y[N];
pair<int, int> tree[N];
struct ask {
	int id, q, x, y;
	ask(int a = 0, int b = 0, int c = 0, int d = 0) {
		id = a, q = b, x = c, y = d;
		if (!x || !y) x = 0, y = 0;
	}
	bool operator < (const ask &b) const {
		return x < b.x;
	}
} Q[N<<2];

namespace BIT {
	int sum[N];
	inline int lowbit(int x) {
		return x & -x;
	}
	void add(int x) {
		for (int i = x; i <= maxn; i += lowbit(i))
			sum[i]++;
	}
	int query(int x) {
		int ans = 0;
		for (int i = x; i >= 1; i -= lowbit(i))
			ans += sum[i];
		return ans;
	}
}
int lsh[N<<3],len;

int main() {
	n = read(), m = read();
	for (int i = 1; i <= n; ++i) {
		tree[i].first = read() + 1, tree[i].second = read() + 1;
		lsh[++len] = tree[i].second;
	}
	int nd = 0;
	for (int i = 1, js = 0; i <= m; ++i) {
		int x = read() + 1, y = read() + 1, X = read() + 1, Y = read() + 1;
		lsh[++len] = Y, lsh[++len] = y - 1;
		if (X and Y)
		Q[++nd] = ask(i, 1, X, Y);
		if (X and (y - 1))
		Q[++nd] = ask(i, -1, X, y - 1);
		if ((x - 1) and Y)
		Q[++nd] = ask(i, -1, x - 1, Y);
		if ((x - 1) and (y - 1))
		Q[++nd] = ask(i, 1, x - 1, y - 1);
	}

	sort(lsh + 1, lsh + 1 + len);
	len = unique(lsh + 1, lsh + 1 + len) - lsh - 1;
	for (int i = 1; i <= n; ++i)
		tree[i].second = lower_bound(lsh + 1, lsh + 1 + len, tree[i].second) - lsh;
	for (int i = 1; i <= nd; ++i) {
		Q[i].y = lower_bound(lsh + 1, lsh + 1 + len, Q[i].y) - lsh;
	}

	sort(tree + 1, tree + 1 + n);
	sort(Q + 1, Q + 1 + nd);

	int js = 1;
	for (int i = 1; i <= nd; ++i) {
		while (tree[js].first <= Q[i].x && js <= n)
			BIT::add(tree[js].second),++js;
		ans[Q[i].id] += Q[i].q * BIT::query(Q[i].y);
	}

	for (int i = 1; i <= m; ++i)
		printf("%d\n", ans[i]);

	return 0;
}