hdu 2642 Stars (二维树状数组)

Stars

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1996    Accepted Submission(s): 848

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
0

思路：用二维树状数组来做，更新单点的时候，要有一个数组记录这个单点是否为亮的。当该点已经亮着则只能进行熄灭的操作，反之只能进行点亮的操作。

最后查询的时候注意这是个二维的，query(x+1,y+1)是算(1,1)到(x+1,y+1)所有亮点的个数。所以要减去两个矩形加上一个矩形的面积才是最后的答案。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 1100
using namespace std;
int tr[MAX_N][MAX_N];
int stars[MAX_N][MAX_N];//存星星的状态
int lowbit(int i)
{
    return i&(-i);
}
void update(int x,int y,int c)
{
    for(int i=x;i<=MAX_N;i+=lowbit(i))
        for(int j=y;j<=MAX_N;j+=lowbit(j))
            tr[i][j]+=c;
}
int query(int x,int y)
{
    int sum=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            sum+=tr[i][j];
    return sum;
}
int main()
{
    int n,x,y;
    scanf("%d",&n);
    while(n--)
    {
        char c[5];
        scanf("%s",c);
        if(c[0]=='B')
        {
            scanf("%d%d",&x,&y);
            if(!stars[x+1][y+1])
            {
                stars[x+1][y+1]=1;
                update(x+1,y+1,1);
            }
        }
        else if(c[0]=='D')
        {
            scanf("%d%d",&x,&y);
            if(stars[x+1][y+1])
            {
                stars[x+1][y+1]=0;
                update(x+1,y+1,-1);
            }
        }
        else
        {
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&x2,&y1,&y2);
            if(x1>x2) swap(x1,x2);//注意题目的输入
            if(y1>y2) swap(y1,y2);
            printf("%d\n",query(x2+1,y2+1)-query(x2+1,y1)-query(x1,y2+1)+query(x1,y1));
        }
    }

}