2018 Blue Bridge Simulation Competition Star Violence in the Sky | 2D Tree Array

On a starry night, Mr. Garantou counted the stars in the sky one by one.

Mr. Garantou ingeniously drew a rectangular coordinate system for the sky, so that all the stars are distributed in the first image. there is n in the sky $n$  stars, he can know the coordinates and brightness of each star.

Now, Mr. Garantou asks himself  $q$  times, each time he asks himself what is the sum of the brightness of the stars in each rectangular area (including the stars on the border).

input format

Enter an integer  $n(1\le n\le 50000)$  represents the number of stars.

next  $n$  lines, each line input three integers $x,y,w(0\le x,y,w\le 2000)$ , expressed at coordinates $(x,y)$  has a brightness $w$  stars. Note that a point may have multiple stars.

Enter an integer  $q(1\le q\le 50000)$ , indicating the number of queries.

next  $q$  lines, each line input four integers x_1, y_1, x_2, y_2x1​, and1​,x2​, and2​, where  (x_1, y_1)(x1​, and1​)  represents the coordinates of the lower left corner of the query rectangle,(x_2, y_2)(x2​, and2​)  represents the coordinates of the upper right corner of the query rectangle,0 \le x_1 \le x_2 \le 20000≤x1​≤x2​≤ 2 0 0 0，0 \le y_1 \le y_2 \le 20000≤y1​≤y2​≤2000。

output format

For each query, output a line with an integer representing the sum of the brightnesses of the stars within the query rectangle.

sample input

5
5 0 6
7 9 7
8 6 13
9 7 1
3 0 19
4
0 8 7 9
0 0 7 10
2 7 10 9
5 4 7 5

Sample output

7
32
8
0

Give the coordinates under the first quadrant and give the values ​​under this coordinates Give us the range of multiple rectangles Find this

Although violence can

But tree-like arrays can also be ~ and faster~

Complexity O(q*log(x)*log(y))

x,y<=2000

Note that the boundaries are counted and the data cannot exist on 0,0

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll tre[2010][2010];
void add(int x,int y,int w)
{
	for(int i=x;i<=2001;i+=i&(-i)){
		for(int j=y;j<=2001;j+=j&(-j)){
			tre[i][j]+=w;
	}
}
ll query(int x,int y){
	ll sum=0;
	for(int i=x;i>0;i-=i&(-i)){
		for(int j=y;j>0;j-=j&(-j)){
			sum+=tre[i][j];			
	}
	return sum;
}
intmain()
{
	int n;
	ios::sync_with_stdio(0);
	cin>>n;
	while(n--){
		int x,y,w;
		cin>>x>>y>>w;
		add(x+1,y+1,w);
	}
	int q;
	cin>>q;
	while(q--){
		int lx,ly,rx,ry;
		cin>>lx>>ly>>rx>>ry;
		
		ll ans = query(rx+1,ry+1)-query(rx+1,ly)-query(lx,ry+1)+query(lx,ly); //Note that the side should be counted in the rectangle
		cout<<ans<<endl;			
	}
	return 0;
}