HDU 5480 Conturbatio（前缀和）-优快云博客

本文链接：https://blog.youkuaiyun.com/xigongdali/article/details/81066702

本文探讨ConturbatioTime中棋盘上的Rook攻击问题，介绍了一种使用前缀和算法来高效判断任意矩形区域是否被所有格子攻击的方法。通过输入测试用例数量、棋盘尺寸、Rook数量及查询次数，算法计算并输出每个查询的判断结果。

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Conturbatio

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1294 Accepted Submission(s): 611

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

   22 2 1 21 11 1 1 22 1 2 22 2 2 11 11 22 1 2 2 
 

Sample Output

   YesNoYes 
   
    Hint 
   
Huge input, scanf recommended.

Source

BestCoder Round #57 (div.2)

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借鉴：https://blog.youkuaiyun.com/u010885899/article/details/48765049

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;
#define MAXN 100005
int row[MAXN];
int column[MAXN];
int main()
{
	int t,i;
	int n,m,k,q,x,y,x1,x2,y1,y2;
	scanf("%d",&t);
	while(t--)
	{
		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]; //算前缀和的关键算法，这样就算好了1 12 123 ... 123..n的所有和，column[i]表示1到i为止的和
		}                                    //2~n==column[n]-column[1],这样所有的和就都算出来了。
		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])  //利用和来进行枚举比较，精妙，这也解释了为什么赋值1的原因
			{
				puts("Yes");
			}
			else
			{
				puts("No");
			}
		}
	}
	return 0;
}