【NJUST5480】Conturbatio-优快云博客

本文链接：https://blog.youkuaiyun.com/better_space/article/details/52289128

本文介绍了一种算法，用于判断棋盘上特定矩形区域是否每个格子都能被至少一个车攻击到。输入包括车的位置及待查询的矩形坐标，通过累积计数优化查询效率。

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Conturbatio

Time Limit: 6000/3000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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Description

There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.

There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?

Input

The first line of the input is a integer $T$, meaning that there are $T$ test cases.

Every test cases begin with four integers $n , m , K , Q$.
$K$ is the number of Rook, $Q$ is the number of queries.

Then $K$ lines follow, each contain two integers $x , y$ describing the coordinate of Rook.

Then $Q$ lines follow, each contain four integers $x1, y1, x2, y2$ describing the left-down and right-up coordinates of query.

$1\leq n , m , K , Q \leq 100,000$.

$1\leq x \leq n , 1 \leq y \leq m$.

$1\leq x1 \leq x2 \leq n , 1 \leq y1 \leq y2 \leq m$.

Output

For every query output "Yes" or "No" as mentioned above.

Sample Input

Sample Output

     
      Yes
No
Yes


      
       
        Hint
       Huge input, scanf recommended.

Hint

hujie

Source

BestCoder Round #57 (div.2)

Related problem

5479 5464 5491 5495 5463

一道比较好的思维题目

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100000+10;
int row[N],col[N];
int main() {
	int T;
	scanf("%d",&T);
	while(T--) {
		memset(row,0,sizeof(row));
		memset(col,0,sizeof(col));
		int n,m,k,q;
		scanf("%d%d%d%d",&n,&m,&k,&q);
		for(int i=0; i<k; i++) {
			int a,b;
			scanf("%d%d",&a,&b);
			row[a]=1;
			col[b]=1;
		}
		for(int i=1; i<=n; i++) {
			row[i]=row[i]+row[i-1];
		}
		for(int i=1; i<=m; i++) {
			col[i]=col[i]+col[i-1];
		}
		while(q--) {
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			if(x2-x1+1==row[x2]-row[x1-1]||y2-y1+1==col[y2]-col[y1-1])
				printf("Yes\n");
			else
				printf("No\n");
		}
	}
	return 0;
}

题目地址：http://acm.hust.edu.cn/vjudge/contest/127335#problem/A