hdoj Conturbatio 5480 （数组模拟+技巧） 好题

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 586    Accepted Submission(s): 274

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

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

Hint

Huge input, scanf recommended.

 

题目大意：有t组测试数据，每组给一个n * m的矩形棋盘格子，棋盘内有很多车（x, y），每个车的攻击范围是这个车所在的行x和列y，共有K辆车，然后有Q次询问，每次询问给一个矩形(四个数描述矩形，左下角和右上角)，问给的矩形的每个格子都能否被车攻击，是就输出“Yes”，否则“No”。

//机智的解法 ，参考大神的 
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100010
using namespace std;
int row[N],col[N];
int main()
{
	int t,i;
	scanf("%d",&t);
	while(t--)
	{
		memset(row,0,sizeof(row));
		memset(col,0,sizeof(col));
		int n,m,k,q;
		scanf("%d%d%d%d",&n,&m,&k,&q);
		for(i=0;i<k;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			row[x]=col[y]=1;//将车的点的横纵坐标标记为1. 
		}
		for(i=1;i<=n;i++)//将车所在的行全标记为1。 
			if(row[i])
				row[i]+=row[i-1];
		for(i=1;i<=m;i++)//将车所在的列全标记为1. 
			if(col[i])
				col[i]+=col[i-1];
		for(i=0;i<q;i++)
		{
			int x1,x2,y1,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			if(row[x2]>=x2-x1+1||col[y2]>=y2-y1+1)
				printf("Yes\n");
			else
				printf("No\n");
		}
	}
	return 0;
}