Stars

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1628    Accepted Submission(s):
683

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

初学树状数组，看了一些文章，算是有了初步的认识。

自己的理解：用二进制表示的方法来实现二分。一个二进制数c[k]表示的是从a[k]开始往前的lowbit(k)个数的和（lowbit(k)为k的最低位，也即将k的除最低位外的所有为置为0）。

更新：

void add(int k,int x)
{
    for(int i=k;i<MAXN;i+=lowbit(i))
        c[i]+=x;
}

lowbit():

int lowbit(int x)   //取最低位
{
    return x&(-x);
}

查询：

int get_sum(int k)
{
    int res=0;
    for(int i=k;i>0;i-=lowbit(i))
        res+=c[i];
    return res;
}

这道题是一道二维树状数组，原理其实也就是这样。

注意：题目中坐标从0开始，可能对一颗star做两次同样的操作。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<stdlib.h>
#include<stack>
#include<vector>
using namespace std;

const int MAXN=1010;
int a[MAXN][MAXN];
bool b[MAXN][MAXN];

int lowbit(int x)
{
    return x&(-x);
}

void modify(int x,int y,int data)
{
    for(int i=x; i<MAXN; i+=lowbit(i))
        for(int j=y; j<MAXN; j+=lowbit(j))
            a[i][j]+=data;
}

int getsum(int x,int y)
{
    int res=0;
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            res+=a[i][j];
    return res;
}

int main()
{
    int n,x,y,x1,y1;
    char str[2];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",str);
        if(str[0]==‘B‘)
        {
            scanf("%d%d",&x,&y);
            x++;
            y++;
            if(b[x][y]) continue;
            modify(x,y,1);
            b[x][y]=1;
        }
        else if(str[0]==‘D‘)
        {
            scanf("%d%d",&x,&y);
            x++;
            y++;
            if(b[x][y]==0) continue;
            modify(x,y,-1);
            b[x][y]=0;
        }
        else
        {
            scanf("%d%d%d%d",&x,&x1,&y,&y1);
            x++;x1++;y++;y1++;
            if(x>x1) swap(x,x1);
            if(y>y1) swap(y,y1);
            int ans=getsum(x1,y1)-getsum(x-1,y1)-getsum(x1,y-1)+getsum(x-1,y-1);
            printf("%d\n",ans);
        }
    }
    return 0;
}