HDU 5480：Conturbatio 前缀和

Conturbatio

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

Total Submission(s): 232    Accepted Submission(s): 108

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.

题意是一个棋盘上有很多车，车可以攻击他所属的一行或一列，包括它自己所在的位置。现在有很多询问，每次询问给定一个棋盘内部的矩形，问矩形内部的所有格子是否都被车攻击到？

对自己的智商感到不断怀疑系列。。。判断中间有没有零，求和啊

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#pragma warning(disable:4996)
using namespace std;

int row[100005];
int column[100005];

int main()
{
	int test,i,j,flag1,flag2;
	int n,m,k,q,x,y,x1,x2,y1,y2;

	scanf("%d",&test);

	while(test--)
	{
		scanf("%d%d%d%d",&n,&m,&k,&q);

		memset(row,0,sizeof(row));
		memset(column,0,sizeof(column));

		for(i=1;i<=k;i++)
		{
			scanf("%d%d",&x,&y);
			row[x]=1;
			column[y]=1;
		}
		for(i=1;i<=n;i++)
		{
			row[i]=row[i]+row[i-1];
		}
		for(i=1;i<=m;i++)
		{
			column[i]=column[i]+column[i-1];
		}
		for(i=1;i<=q;i++)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			if(x2-x1+1==row[x2]-row[x1-1]||y2-y1+1==column[y2]-column[y1-1])
			{
				puts("Yes");
			}
			else
			{
				puts("No");
			}
		}
	}
	return 0;
}

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