poj1195(二维树状数组)

Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 12554		Accepted: 5787

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

Source

 

本题需要反复修改、查询区间的值，可以考虑用线段树或树状数组，又由于所操作区间是矩阵，故用二维的；

二维树状数组的实现比二维线段树简单，且时空复杂度相同，但前者常数较小，故树状数组占优，此处不详细说明二维树状数组的原理

 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include <vector>
using namespace std;
#define MY_MAX 1100
int C[MY_MAX][MY_MAX];
int Lowbit[MY_MAX];
int s;

void Add( int y, int x,int a)
{
 while( y <= s )
 {
  int tmpx = x;
  while( tmpx <= s )
  {
   C[y][tmpx] += a;
   tmpx += Lowbit[tmpx];
  }
  y += Lowbit[y];
 }
}

int QuerySum( int y, int x)
//查询第1行到第y行，第1列到第x列的和
{
 int nSum = 0;
 while( y > 0 )
 {
  int tmpx = x;
  while( tmpx > 0)
  {
   nSum += C[y][tmpx];
   tmpx -= Lowbit[tmpx];
  }
  y -= Lowbit[y];
 }
 return nSum;
}

int main()
{
 int cmd; int x,y,a,l,b,r,t;
 int i,j,k;
 int n1,n2;
 for( i = 1;i <= MY_MAX;i ++ )
  Lowbit[i] = i & ( i ^(i - 1));
 while( true)
 {
  scanf("%d",&cmd);
  switch( cmd)
  {
  case 0:
   scanf("%d",& s);
   memset( C,0,sizeof(C));
   break;
  case 1:
   scanf("%d%d%d",&x ,&y,&a);
   Add( y + 1, x + 1, a);
   break;
  case 2:
   scanf("%d%d%d%d",&l , &b, &r,&t);
   int n1,n2,n3,n4;
   l ++; b++; r ++; t ++;
   //输出小矩形内的元素和
   printf("%d\n",QuerySum(t,r) +QuerySum(b-1,l-1) - QuerySum(t,l-1) - QuerySum(b-1,r));
   break;
  case 3: return 0;
  }
 }
}