HDU5480： Conturbatio

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 , meaning that there are  test cases. 

Every test cases begin with four integers . 
 is the number of Rook,  is the number of queries. 

Then  lines follow, each contain two integers  describing the coordinate of Rook. 

Then  lines follow, each contain four integers  describing the left-down and right-up coordinates of query. 

. 

. 

. 

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

给一个矩阵（给的是左上角和右下角坐标）  是否能够被车全部攻击到
只要矩阵每行都有一个车  或者  每列有一个车就可以

记录哪行哪列有车

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int x[100001];
int y[100001];
int main()
{
	int t;
	int n,m,k,q;
	scanf("%d",&t);
	while(t--)
	{
		memset(x,0,sizeof(x));
		memset(y,0,sizeof(y));
		scanf("%d%d%d%d",&n,&m,&k,&q);
		int dx,dy;
		while(k--)
		{
			scanf("%d%d",&dx,&dy); 
			x[dx]=1;
			y[dy]=1;
		}
	   for(int i=1;i<=n;i++)
	   {
	   	x[i]+=x[i-1];//这里我有特意画了个图理解
	   }
	   for(int i=0;i<=m;i++)
	   {
	   	y[i]+=y[i-1];
	   }
	   int x1,y1,x2,y2;
	   while(q--)
	   {
	   	scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
	   	if(x[x2]-x[x1-1]==x2-x1+1||y[y2]-y[y1-1]==y2-y1+1)
	   	printf("Yes\n");
		   else
		   printf("No\n"); 
	   }
	 } 
	return 0;
}