hdoj5480 Conturbatio(思维)

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
有一个n*m的棋盘，有k个车（和象棋里的车一样），有q次询问，问车是否能将给出的矩形的所有格子打到，给出车的坐标，矩形左下角的坐标和右上角的坐标.cow[i]表示以第i行结尾的矩形，col[i]表示以第i列结尾的矩形.
代码如下：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int cow[100010];
int col[100010];
int main()
{
	int n,m,k,q;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d%d",&n,&m,&k,&q);
		memset(cow,0,sizeof(cow));
		memset(col,0,sizeof(col));
		int i;int x,y;
		for(i=0;i<k;i++)
		{
		scanf("%d%d",&x,&y);
		cow[x]=col[y]=1;	
		}
	
		for(i=1;i<=n;i++)
		{
			if(cow[i])
			cow[i]+=cow[i-1];
		}
		for(i=1;i<=m;i++)
		{
			if(col[i])
			col[i]+=col[i-1];
		}
		int x1,x2,y1,y2;
		for(i=1;i<=q;i++)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			if(cow[x2]>=x2-x1+1||col[y2]>=y2-y1+1)
			printf("Yes\n");
			else
			printf("No\n");
		}
	}
}