HDU 2642 Stars (二维树状数组)_0521fenfei-优快云博客

本文详细介绍了Yifenfei如何通过计算天空中明亮星星的数量来简化星河计数的问题。通过一系列操作，如标记星星的亮度变化、区域查询等，实现对特定区域内明亮星星数量的快速准确计算。

Stars

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 711 Accepted Submission(s): 309

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

Author

teddy

Source

百万秦关终属楚

Recommend

teddy

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=1005;
int c[maxn][maxn];
bool f[maxn][maxn];
inline int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int val)
{
    int yt;
    f[x][y]=(val==1?1:0);
    while(x<=maxn)
    {
        yt=y;
        while(yt<=maxn)
        {
            c[x][yt]+=val;
            yt+=lowbit(yt);
        }
        x+=lowbit(x);
    }
}
int sum(int x,int y)
{
    int s=0,yt;
    while(x>0)
    {
        yt=y;
        while(yt>0)
        {
            s+=c[x][yt];
            yt-=lowbit(yt);
        }
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int m,a,b,d,e;
    char op[2];
    while(~scanf("%d",&m))
    {
        memset(c,0,sizeof(c));
        memset(f,0,sizeof(f));
        while(m--)
        {
            scanf("%s%d%d",op,&a,&b);
            a++,b++;
            if(op[0]=='B')
            {
                if(f[a][b])continue;
                update(a,b,1);
            }
            else if(op[0]=='D')
            {
                if(!f[a][b])continue;
                update(a,b,-1);
            }
            else
            {
                scanf("%d%d",&e,&d);
                e++,d++;
                int ix=min(a,b),ax=max(a,b),iy=min(e,d),ay=max(e,d);
                printf("%d\n",sum(ax,ay)+sum(ix-1,iy-1)-sum(ix-1,ay)-sum(ax,iy-1));
            }
        }
    }
    return 0;
}