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.

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
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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posted on 2015-09-27 09:22  光速小子  阅读(177)  评论(0编辑  收藏  举报

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