分析

一个非常显然的单调队列+线段树的常规题
当然，扫描线一样一样的

但因为以前是直接动态开点的，所以这次打个离散化版本

\(Code\)

#include<cstdio>
#include<algorithm>
#include<iostream>
#define LL long long
#define ls (k << 1)
#define rs (ls | 1)
using namespace std;

const int N = 10004;
int n, W, H, X[N], Y[N], T, d[N];
LL ans, mx[N << 2], tag[N << 2];
struct point{int x, y, v;}a[N];

inline bool cmp1(point a, point b){return a.x < b.x;}
inline bool cmp2(point a, point b){return a.y < b.y;}

void build(int l, int r, int k)
{
	mx[k] = tag[k] = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(l, mid, ls), build(mid + 1, r, rs);
}
void update(int l, int r, int k, int x, int y, int v)
{
	if (x <= l && r <= y) return void(tag[k] += v);
	int mid = (l + r) >> 1;
	if (x <= mid) update(l, mid, ls, x, y, v);
	if (y > mid) update(mid + 1, r, rs, x, y, v);
	mx[k] = max(mx[ls] + tag[ls], mx[rs] + tag[rs]);
}

inline int search_up(int y)
{
	int l = 1, r = n, mid, ret;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (Y[mid] >= y) ret = mid, r = mid - 1;
		else l = mid + 1;
	}
	return ret;
}

int main()
{
	scanf("%d", &T);
	for(; T; T--)
	{
		scanf("%d%d%d", &n, &W, &H);
		ans = 0, build(1, n, 1);
		for(register int i = 1; i <= n; i++) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].v);
		sort(a + 1, a + n + 1, cmp2);
		for(register int i = 1; i <= n; i++) Y[i] = a[i].y, a[i].y = i;
		sort(a + 1, a + n + 1, cmp1);
		for(register int i = 1; i <= n; i++) X[i] = a[i].x, a[i].x = i;
		int h = 1, t = 0;
		for(register int i = 1; i <= n; i++)
		{
			while (h <= t && X[a[i].x] - X[a[d[h]].x] >= W) 
				update(1, n, 1, search_up(Y[a[d[h]].y] - H + 1), a[d[h]].y, -a[d[h]].v), ++h;
			d[++t] = i, update(1, n, 1, search_up(Y[a[i].y] - H + 1), a[i].y, a[i].v);
			ans = max(ans, mx[1] + tag[1]);
		}
		printf("%lld\n", ans);
	}
}

Luogu P1502.窗口的星星

原文：https://www.cnblogs.com/leiyuanze/p/14280319.html