Mobile Phone（二维树状数组）

Mobile Phone

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Source : IOI 2001Time limit : 5 secMemory limit : 32 M

Submitted : 1040, Accepted : 310

PROBLEM

 

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 AND OUTPUT

 

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

 

Instruction

Parameters

Meaning

0

S

Initialize the matrix size to S´S containing all zeros. This instruction is given only once and it will be the first instruction.

1

X Y A

Add A to the number of active phones in table square (X, Y). Amay be positive or negative.

2

L B R T

Query the current sum of numbers of active mobile phones  in squares (X,Y), where L £ X £ R, B £ Y £ T

3

 

Terminate program. This instruction is given only once and it will be the last instruction.

 

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.

 

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.

 

There are multiple test cases, process to the end of file.

CONSTRAINTS

 

0 £ V £ 2¹⁵ –1  (= 32767)

Table size

S*S

1´1 £ S´S £ 1024´1024

Cell value V at any time

V

0 £ V £ 2¹⁵ –1  (= 32767)

Update amount

A

- 2¹⁵ £ A £ 2¹⁵–1  (= 32767)

No of instructions in input

U

3 £ U £ 60002

Maximum number of phones in the whole table

M

M= 2³⁰

 

Sample Input

0 41 1 2 32 0 0 2 21 1 1 21 1 2 -12 1 1 2 33

Sample Output

34

这是一个较为经典的二维树状数组的题目，其实二维相较于一维也就是处理的时候多了一维

例如一维的add函数由

void add(int x,LL Add){for(int i = x; i <= S; i = i+lowbit(i)){c[i] += Add;}}

更新为二维的

void add(int x,int y,LL Add){for(int i = x; i <= S; i = i+lowbit(i)){for(int j = y; j <= S; j = j+lowbit(j)){Map[i][j] += Add;}}}

其他的与之类似，比如sum函数，其他的就都一样了

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>using namespace std;typedef long long LL;const int INF = 0x3f3f3f3f;const int maxn = 1030;int S,X,Y,ADD;int X1,X2,Y1,Y2;LL Map[maxn][maxn];int lowbit(int x){return (x & -x);}void add(int x,int y,LL Add){for(int i = x; i <= S; i = i+lowbit(i)){for(int j = y; j <= S; j = j+lowbit(j)){Map[i][j] += Add;}}}LL sum(int x,int y){LL ans = 0;for(int i = x; i > 0; i = i-lowbit(i))for(int j = y; j > 0; j = j-lowbit(j)){ans += Map[i][j];}return ans;}int main(){int ch;while(~scanf("%d %d",&ch,&S)){memset(Map,0,sizeof(Map));while(scanf("%d",&ch) && ch != 3){if(ch == 1){scanf("%d %d %d",&X,&Y,&ADD);add(X+1,Y+1,ADD);}else{scanf("%d %d %d %d",&X1,&Y1,&X2,&Y2);printf("%lld\n",sum(X2+1,Y2+1)-sum(X1,Y2+1)-sum(X2+1,Y1)+sum(X1,Y1));}}}return 0;}