hdoj5480 Conturbatio(思维)

本文介绍了一种算法,用于判断棋盘上的多个车是否能够覆盖指定矩形区域内的所有格子。通过预处理行和列的攻击状态,该算法能够在O(1)时间内回答每个查询。
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.

1n,m,K,Q100,000.

1xn,1ym.

1x1x2n,1y1y2m.
 

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");
		}
	}
}


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