Counting Rectangles

题面翻译

分别给定 $n$ 个正整数 $h_i,w_i$ ，有 $q$ 次询问。

对于每次询问，输入四个整数 $h_s,w_s,h_b,w_b$ ，输出满足 $h_i \in (h_s,h_b)$ , $w_i\in(w_s,w_b)$ 的 $\sum h_i \cdot w_i$

输入格式

The first line of the input contains an integer $t$ ( $1 \leq t \leq 100$ ) — the number of test cases.

The first line of each test case two integers $n, q$ ( $1 \leq n \leq 10^5$ ; $1 \leq q \leq 10^5$ ) — the number of rectangles you own and the number of queries.

Then $n$ lines follow, each containing two integers $h_i, w_i$ ( $1 \leq h_i, w_i \leq 1000$ ) — the height and width of the $i$ -th rectangle.

Then $q$ lines follow, each containing four integers $h_s, w_s, h_b, w_b$ ( $1 \leq h_s < h_b,\ w_s < w_b \leq 1000$ ) — the description of each query.

The sum of $q$ over all test cases does not exceed $10^5$ , and the sum of $n$ over all test cases does not exceed $10^5$ .

输出格式

For each test case, output $q$ lines, the $i$ -th line containing the answer to the $i$ -th query.

样例 #1

样例输入 #1

3
2 1
2 3
3 2
1 1 3 4
5 5
1 1
2 2
3 3
4 4
5 5
3 3 6 6
2 1 4 5
1 1 2 10
1 1 100 100
1 1 3 3
3 1
999 999
999 999
999 998
1 1 1000 1000

样例输出 #1

提示

In the first test case, there is only one query. We need to find the sum of areas of all rectangles that can fit a $1 \times 1$ rectangle inside of it and fit into a $3 \times 4$ rectangle.

Only the $2 \times 3$ rectangle works, because $1 < 2$ (comparing heights) and $1 < 3$ (comparing widths), so the $1 \times 1$ rectangle fits inside, and $2 < 3$ (comparing heights) and $3 < 4$ (comparing widths), so it fits inside the $3 \times 4$ rectangle. The $3 \times 2$ rectangle is too tall to fit in a $3 \times 4$ rectangle. The total area is $2 \cdot 3 = 6$ .

分析

我开始又在想二分，蚌埠住了，后来发现不是单调性，没法二分，然后我就不会做了，~~你怎么这么快啊~~，于是发现这其实是一个二维前缀和的板子，没法再简单了，就当是复习一下二维前缀和吧。。。这里要注意要把输入的前区间 $++$ ,后区间 $--$ 。因为是开区间，欧了就。。。

代码

#include <bits/stdc++.h>

#define ll long long
#define x first
#define y second

using namespace std;
const int N=1010;
ll s[N][N];
int n,q;
void solve(){
	cin>>n>>q;
	memset(s,0,sizeof s);
	for(int i=1;i<=n;i++){
		ll h,w;
		cin>>h>>w;
		s[h][w]+=h*w;
	}
	for(int i=1;i<=1000;i++){
		for(int j=1;j<=1000;j++){
			s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1]; 
		} 
	}
	while(q--){
		int l1,r1,l2,r2;
		cin>>l1>>r1>>l2>>r2;
		l1++;r1++;l2--;r2--; 
		cout<<s[l2][r2]-s[l1-1][r2]-s[l2][r1-1]+s[l1-1][r1-1]<<"\n";
	}
//	while(q--){
//		int l1,r1,l2,r2;
//		cin>>l1>>r1>>l2>>r2;
//		l1++;r1++;l2--;r2--;
//		cout<<s[l1][r2]-s[l1-1][r2]-s[l2][r1-1]+s[l1-1][r1-1]<<"\n";
//	}
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int t;
	cin>>t;
	while(t--){
		solve();
		
	}
	return 0;
}

CF1722E Counting Rectangles(div4)