BestCoder Round #57 (div.2)1002

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 43    Accepted Submission(s): 21

Problem 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≤n,m,K,Q≤100,000.

1≤x≤n,1≤y≤m.

1≤x1≤x2≤n,1≤y1≤y2≤m.

 

Output

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

 

Sample Input

22 2 1 21 11 1 1 22 1 2 22 2 2 11 11 22 1 2 2

 

Sample Output

YesNoYes
Hint
Huge input, scanf recommended.
 

 

其实也是一个简单题；

这个题目最大的问题是在询问，每个询问要在O(logn)或O(1)时间完成

一开始以为是树状数组，当后来发现数组开不了那么大

询问那个矩形内是否会受到攻击，就是看它的每一行有一个车或它的每一列有一个车。

如果有就输出Yes，否者输出No

这样就很简单了，直接用两个数组标记那些行和列有车，然后得到其前缀和（用来判断它的每一行是否有一个车或每一列是否有一个车）

hang[i]表示前i行车的个数

lie[i]表示前i列车的个数

对于x1,y1,x2,y2这个矩形

只要判断hang[y2]-hang[y1-1] == y2-y1+1 和 lie[x2]-lie[x1-1] == x2-x1+1有一个成立就输出Yes否者输出No