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

2
2 2 1 2
1 1
1 1 1 2
2 1 2 2
2 2 2 1
1 1
1 2
2 1 2 2

Sample Output

Yes
No
Yes

HintHuge 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