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): 633    Accepted Submission(s): 269

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
   
   
   
   

 

话说，这是最最裸的二维树状数组，只要掌握了基本的树状数组，在二维的数组里每次更新或是查询的时候多一步操作的可以了~

#include <cstdio>
#include <iostream>
#include <string.h>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int map[1005][1005];
int visit[1005][1005];
int lowbit( int t )
{
    return t&(-t) ;   
}

int query( int x , int y )
{
    int sum = 0;
    int ry ;
    while( x > 0 ){
           ry = y;
           while( ry > 0 ) {
                  sum += map[x][ry];
                  ry -= lowbit( ry );
           }
           x -= lowbit( x );
    }
    return sum;
}

void update( int x ,  int y , int val )
{    
     int ry;
     while( x < 1005 )
     {
            ry = y;
            while( ry < 1005 ){
                   map[x][ry] += val;
                   ry += lowbit( ry );
            }
            x += lowbit( x );
     }
}

int main( )
{
    int n;
    while( scanf("%d",&n) != EOF )
    {
           getchar();
           memset( map , 0 , sizeof(map) );
           memset( visit , 0 , sizeof(visit) );
           char s[10];
           int x1,x2,y1,y2;
           int maxx , maxy , minx, miny;
           while( n-- ){
                  scanf("%s",s);
                 if( s[0] == 'B' ){
                     scanf("%d%d",&x1,&y1);
                     if( visit[x1+1][y1+1] == 0 ){
                         visit[x1+1][y1+1] = 1;
                         update( x1+1 , y1+1 , 1 );
                     }
                     
                 }
                 else if( s[0] == 'D' ){
                      scanf("%d%d",&x1,&y1);
                      if( visit[x1+1][y1+1] == 1 ){
                          update( x1+1 , y1+1 , -1 );
                          visit[x1+1][y1+1] = 0;
                      }
                 }
                 else if( s[0] == 'Q' ){
                      scanf("%d%d%d%d",&x1,&x2,&y1,&y2);
                      maxx = max( x1 , x2 );
                      maxy = max( y1 , y2 );
                      minx = min( x1 , x2 );
                      miny = min( y1 , y2 );
                      int ret1 , ret2 , ret3 , ret4;
                      ret1 = query(maxx+1 , maxy+1);
                      ret2 = query( maxx+1 , miny );
                      ret3 = query( minx , maxy+1 );
                      ret4 = query( minx , miny );
                      printf("%d\n", ret1 + ret4 - ret3 - ret2 );     
                 }
           }
    }
    return 0;
}