POJ 1195

Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 13774		Accepted: 6393

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

题目的大意是在一个二维的数组里，更新某些值，询问（l,b,r,t）之内的值的和。有4种操作：0：输入一个S，建立一个S*S的矩阵；1：将A值加到相应的（x,y）里；2：求值。

注意范围。对于二维的add(x, y)，c[x][y]记录的是以原点和(x, y)做的矩形中的值的和，比如说c[x][y] = ∑( c[ai][bi] )，c[ai][bi]是其中的子矩形的值的和。

 1 /*

 2 POJ 1195 Mobile phones

 3 二维树状数组

 4 */

 5 #include<stdio.h>

 6 #include<iostream>

 7 #include<string.h>

 8 using namespace std;

 9 #define N 1030

10 int c[N][N];

11 int n;

12 int lowbit(int x)

13 {

14     return x & (-x);

15 }

16 void add(int x, int y, int val)

17 {

18     for(int i=x; i<=n; i+=lowbit(i))

19         for(int j=y; j<=n; j+=lowbit(j))

20             c[i][j]+=val;

21 }

22 int getsum(int x, int y)

23 {

24     int s=0;

25     for(int i=x; i>0; i-=lowbit(i))

26         for(int j=y; j>0; j-=lowbit(j))

27             s+=c[i][j];

28     return s;

29 }

30 int main()

31 {

32     //freopen("1195.txt","r",stdin);

33     int X,Y,A,L,B,R,T;

34     int a;

35     while(scanf("%d",&a)!=EOF)

36     {

37         if(a==0)

38         {

39             scanf("%d",&n);

40             memset(c,0,sizeof(c));

41         }

42         if(a==1)

43         {

44             scanf("%d %d %d",&X,&Y,&A);

45             add(X+1,Y+1,A);

46         }

47         if(a==2)

48         {

49             scanf("%d %d %d %d",&L,&B,&R,&T);

50             printf("%d\n",getsum(R+1,T+1)+getsum(L,B)-getsum(L,T+1)-getsum(R+1,B));

51         }

52         if(a==3) break;

53     }

54 

55     return 0;

56 }