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最新推荐文章于 2021-05-30 16:17:32 发布

敢于挑战才会成功

最新推荐文章于 2021-05-30 16:17:32 发布

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分类专栏：树状数组/线段树

本文链接：https://blog.youkuaiyun.com/wangxianse/article/details/49934307

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本文介绍了一个使用二维树状数组解决特定星空计数问题的算法案例。该算法能够高效地处理星星亮起和熄灭的操作，并快速查询指定区域内亮星的数量。代码实现了输入处理、树状数组更新及查询等功能。

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Problem Description
Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y" where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.

There is only one case.

Input
The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.

Output
For each query,output the number of bright stars in one line.

Sample Input
5
B 581 145
B 581 145
Q 0 600 0 200
D 581 145
Q 0 600 0 200

Sample Output
1

二维树状数组

代码：

#include <iostream>
#include <algorithm>
#include <cstring> 
using namespace std;
int a[1005][1005],c[1005][1005];
int lowbit(int x)
{
	return x&(-x);
}
int Sum(int i,int j)
{   int sum=0,k,p;
	for(k=i;k>0;k=k-lowbit(k))
	  for(p=j;p>0;p=p-lowbit(p))
	    sum+=c[k][p];
	return sum;
}
void change(int i,int j,int x)
{       int k,p;
		for(k=i;k<=1004;k=k+lowbit(k))
	      for(p=j;p<=1004;p=p+lowbit(p))
	          c[k][p]+=x;
}
int main()
{
	int i,j,k,n,m,x,y;
	char ch;
	    while(cin>>n)
	    {
	     m=n;
		 memset(a,0,sizeof(a));
		 memset(c,0,sizeof(c));
		 while(n--)
		  {
			cin>>ch;
			if(ch=='B'||ch=='D')
			{
				cin>>x>>y;
				x++;y++;
				if(ch=='B'&&a[x][y]==0)
				    {
					  change(x,y,1); 
					  a[x][y]=1;
				    }
			    
                if(ch=='D'&&a[x][y]==1)
                    { 
					  change(x,y,-1);
					  a[x][y]=0;
                    }
		    }
            else 
		    {	int x1,x2,y1,y2;
		        cin>>x1>>x2>>y1>>y2;
				if(x1>x2)
			        swap(x1,x2);
			
				if(y1>y2)
					swap(y1,y2);
				cout<<Sum(x2+1,y2+1)+Sum(x1,y1)-Sum(x1,y2+1)-Sum(x2+1,y1)<<endl;
		    }
		   
		  } 
	    }
	return 0;
}