Poj1195 Mobile phones

Poj1195 Mobile phones

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 has es 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 particle, 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 shocould 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 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

3

4

We can use a two-dimensional tree array to solve this problem. We feel that a two-dimensional array is in the square form of a one-dimensional array. we can add a dimension, and then convert it into two loops. It is not difficult to understand this question. Note that the coordinates of the rectangle must be added with 1 (that is, all coordinates are moved to the right). In this way, getsum (f, g) is used to calculate the area of the rectangle) + getsum (D-1, E-1)-getsum (f, E-1)-getsum (D-1, g) This formula avoids the occurrence of D-1 = 0 or E-1 = 0 because lowbit (0) = 0, this will be in an infinite loop, and the size of maxn cannot be opened to 2 ^ k, such as 2048, because when I reaches 2048, array a [2048] [2048] may be out of range. You can also open the array to a value slightly greater than maxn.

 

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          #includeusing namespace std;#define maxn 2000int b[maxn][maxn];int lowbit(int x){return x&(-x);}void update(int x,int y,int num){int i,j;for(i=x;i<=maxn;i+=lowbit(i)){for(j=y;j<=maxn;j+=lowbit(j)){b[i][j]+=num;}}}int getsum(int x,int y){int num=0,i,j;for(i=x;i>0;i-=lowbit(i)){for(j=y;j>0;j-=lowbit(j)){num+=b[i][j];}}return num;}int main(){int n,m,i,j,c,d,f,g,e;while(scanf("%d%d",&m,&n)!=EOF){memset(b,0,sizeof(b));while(1){scanf("%d",&c);if(c==3)break;if(c==1){ scanf("%d%d%d",&d,&e,&f); d++;e++;//f++; update(d,e,f); } else if(c==2){ scanf("%d%d%d%d",&d,&e,&f,&g); d++;e++;f++;g++; printf("%d\n",getsum(f,g)+getsum(d-1,e-1)-getsum(f,e-1)-getsum(d-1,g)); }}}}