【原】 POJ 1195 Mobile phones 2D树状数组 解题报告

 

http://poj.org/problem?id=1195

 

方法：
简单的2D树状数组的应用

树状数组BIT形状很像二项树，适用于对经常改变的数组快速求得区间和。
采用树状数组tree[N]的话，每次调整与求和的复杂度为O(logN)，效率大大提高。

介绍BIT的好文
http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees
http://blog.sina.com.cn/s/blog_49c5866c0100f4l7.html

 

Description

Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30

Output

Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 4

1 1 2 3

2 0 0 2 2

1 1 1 2

1 1 2 -1

2 1 1 2 3

3

Sample Output

3

4

 

   1: #include <stdio.h>

   2:  

   3: const int N = 1026 ;

   4:  

   5: int tree[N][N] ;

   6:  

   7: int GetSum( int x, int y )

   8: {

   9:     int y1 ;

  10:     int sum = 0 ;

  11:     while( x>0 )

  12:     {

  13:         y1 = y ;

  14:         while( y1>0 )

  15:         {

  16:             sum += tree[x][y1] ;

  17:             y1 -= ( y1 & -y1 ) ;

  18:         }

  19:         x -= ( x & -x ) ;

  20:     }

  21:  

  22:     return sum ;

  23: }

  24:  

  25: void AddVal( int n, int x, int y, int val )

  26: {

  27:     int y1 ;

  28:     while( x<=n )

  29:     {

  30:         y1 = y ;

  31:         while( y1<=n )

  32:         {

  33:             tree[x][y1] += val ;

  34:             y1 += ( y1 & -y1 ) ;

  35:         }

  36:         x += ( x & -x ) ;

  37:     }

  38: }

  39:  

  40: void run1195()

  41: {

  42:     int n ;

  43:     int instruction ;

  44:     int x1,y1,x2,y2,val ;

  45:  

  46:     scanf( "%d%d" , &instruction, &n ) ;

  47:  

  48:     while( scanf( "%d", &instruction ) && instruction!=3 )

  49:     {

  50:         if( instruction == 1 )

  51:         {

  52:             scanf( "%d%d%d", &x1,&y1,&val ) ;

  53:             ++x1 ; ++y1 ;  //避免idx为0

  54:             AddVal( n, x1, y1, val ) ;

  55:         }

  56:         else if( instruction == 2 )

  57:         {

  58:             scanf( "%d%d%d%d", &x1,&y1,&x2,&y2 ) ;

  59:             ++x1 ; ++y1 ; ++x2 ; ++y2 ; //避免idx为0

  60:             printf( "%d\n", GetSum(x2,y2)-GetSum(x1-1,y2)-GetSum(x2,y1-1)+GetSum(x1-1,y1-1) ) ;

  61:         }

  62:     }

  63: }