BZOJ1935或洛谷2163 [SHOI2007]园丁的烦恼

BZOJ原题链接

洛谷原题链接

很容易想到二维前缀和。
设(S[i][j])表示矩阵((0, 0)(i, j))内树木的棵数，则询问的矩形为((x, y)(xx, yy))时，答案为(S[xx][yy] - S[x - 1][yy] - S[xx][y - 1] + S[x - 1][y - 1])。
但这题坐标极大，显然不能直接求。
对(x,y)都进行离散化，然后我们考虑求询问。
将询问的矩阵拆成二维前缀和计算形式的四个矩阵，这样就可以用扫描线快速求矩阵，并统计答案即可。
这里我是用树状数组来维护的。
因为偷懒就用了(vector)

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 5e5 + 10;
struct dd {
	int x, y, xx, yy;
};
dd a[N];
int tr_x[N], tr_y[N], ls_x[N << 2], ls_y[N << 2], C[N], an[N], nl, ml, xl, yl;
vector<int>X[N], q_1[N], q_2[N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline int lowbit(int x) { return x & -x; }
inline void add(int x)
{
	for (; x <= ml; x += lowbit(x))
		C[x]++;
}
inline int ask(int x)
{
	int s = 0;
	for (; x; x -= lowbit(x))
		s += C[x];
	return s;
}
inline int BSX(int x)
{
	int l = 1, r = nl, mid;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (!(ls_x[mid] ^ x))
			return mid;
		ls_x[mid] > x ? r = mid - 1 : l = mid + 1;
	}
	return 0;
}
inline int BSY(int x)
{
	int l = 1, r = ml, mid;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (!(ls_y[mid] ^ x))
			return mid;
		ls_y[mid] > x ? r = mid - 1 : l = mid + 1;
	}
	return 0;
}
int main()
{
	int i, j, n, m, L;
	n = re();
	m = re();
	for (i = 1; i <= n; i++)
	{
		tr_x[i] = re() + 1;
		tr_y[i] = re() + 1;
		ls_x[++xl] = tr_x[i];
		ls_y[++yl] = tr_y[i];
	}
	for (i = 1; i <= m; i++)
	{
		a[i].x = re();
		a[i].y = re();
		a[i].xx = re() + 1;
		a[i].yy = re() + 1;
		ls_x[++xl] = a[i].x;
		ls_x[++xl] = a[i].xx;
		ls_y[++yl] = a[i].y;
		ls_y[++yl] = a[i].yy;
	}
	sort(ls_x + 1, ls_x + xl + 1);
	sort(ls_y + 1, ls_y + yl + 1);
	ls_x[xl + 1] = ls_y[yl + 1] = -1;
	for (i = 1; i <= xl; i++)
		if (ls_x[i] ^ ls_x[i + 1])
			ls_x[++nl] = ls_x[i];
	for (i = 1; i <= yl; i++)
		if (ls_y[i] ^ ls_y[i + 1])
			ls_y[++ml] = ls_y[i];
	for (i = 1; i <= n; i++)
		X[BSX(tr_x[i])].push_back(BSY(tr_y[i]));
	for (i = 1; i <= m; i++)
	{
		a[i].y = BSY(a[i].y);
		a[i].yy = BSY(a[i].yy);
		q_1[BSX(a[i].x)].push_back(i);
		q_2[BSX(a[i].xx)].push_back(i);
	}
	for (i = 1; i <= nl; i++)
	{
		for (j = 0, L = X[i].size(); j < L; j++)
			add(X[i][j]);
		for (j = 0, L = q_1[i].size(); j < L; j++)
			an[q_1[i][j]] += ask(a[q_1[i][j]].y) - ask(a[q_1[i][j]].yy);
		for (j = 0, L = q_2[i].size(); j < L; j++)
			an[q_2[i][j]] += ask(a[q_2[i][j]].yy) - ask(a[q_2[i][j]].y);
	}
	for (i = 1; i <= m; i++)
		printf("%d
", an[i]);
	return 0;
}