POJ 1195 Mobile phones - 二维树状数组/线段树

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 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

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

二维树状数组的模板题

N.B.循环找C数组时切记需要将x，y的值赋给循环变量，否则会出错

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn=1500;

int c[maxn][maxn];
int n;

inline int lowbit(int x)
{
return x&(-x);
}
void change(int x,int y,int val)
{
for(int i=x;i<=n;i+=lowbit(i))
for(int j=y;j<=n;j+=lowbit(j)) //N.B.
c[i][j]+=val;
}
int sum(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_;
}
inline void add()
{
int x,y,val;
scanf("%d%d%d",&x,&y,&val);
x++,y++;
change(x,y,val);
}
inline void cal()
{
int x1,x2,y1,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
x1++;y1++;x2++;y2++;
cout<<sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)<<endl;
}
int main()
{
int order;
while(~scanf("%d",&order))
{
switch(order)
{
case 0:scanf("%d",&n);break;
case 1:add();break;
case 2:cal();break;
case 3:return 0;
}
}
}