hdu5480 Conturbatio（思维）

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 975    Accepted Submission(s): 439

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

Hint
Huge input, scanf recommended.
 

 

题意：给你n*m的方格，给你k个车，其中车可以攻击他所属的一行或一列，包括它自己所在的位置,.再给你q行，每行4个整数x1 , y1 , x2 , y2x1,y1,x2,y2，表示询问的矩形的左下角与右上角的坐标。

可以开两个数组，一个表示行，一个表示列，一个车所在的行列记录下来。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 100005
int r[N],c[N];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(r,0,sizeof(r));
		memset(c,0,sizeof(c)); 
		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);
			r[a]=1;
			c[b]=1;
		}
		for(int i=2;i<=n;i++)
		{
			r[i]+=r[i-1];
		}
		for(int i=2;i<=m;i++)
		{
			c[i]+=c[i-1];
		}
		for(int i=0;i<q;i++)
		{
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
		    if(r[x2]-r[x1-1]==x2-x1+1||c[y2]-c[y1-1]==y2-y1+1)//如果两个条件都不满足，说明至少有一行一列没有车，那这一个行列所对应的值也就不能被攻击到 
			    printf("Yes\n");
			else printf("No\n");
		}
	}
	return 0;
}