poj1195——Mobile phones（二维树状数组）-白红宇

poj1195——Mobile phones（二维树状数组）

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发布时间：2019-05-10

本文共 2743 字，大约阅读时间需要 9 分钟。

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

Sample Output

这道题好像也能用线段树，不过树状数组更容易理解，二维的操作和普通的几乎一毛一样

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define INF 0x3f3f3f3f#define MAXN 2000#define Mod 10001using namespace std;int c[MAXN][MAXN],n;int lowbit(int k){ return k&-k;}int read(int x,int y){ int sum=0; for(int i=x;i>0;i-=lowbit(i)) for(int j=y;j>0;j-=lowbit(j)) sum+=c[i][j]; return sum;}void add(int x,int y,int a){ for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)) c[i][j]+=a;}int main(){ int op; while(~scanf("%d%d",&op,&n)) { memset(c,0,sizeof(c)); while(1) { scanf("%d",&op); if(op==1) { int x,y,a; scanf("%d%d%d",&x,&y,&a); add(x+1,y+1,a); } if(op==2) { int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); x1++,y1++,x2++,y2++; printf("%d\n",read(x2,y2)-read(x2,y1-1)-read(x1-1,y2)+read(x1-1,y1-1)); } if(op==3) break; } } return 0;}

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