POJ1195 Mobile phones 【二维树状数组】

Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 14288		Accepted: 6642

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

Sample Output

3
4

【问题描写叙述】

如果第四代移动电话的收发站是这样执行。整个区域被切割成非常小的方格。全部的方格组成了一个S*S的矩阵，行和列从0~S-1编号。每一个小方格都包括一个收发站。每一个方格内的手机的移动电话数量能够不断改变，由于手机用户在各个方格之间移动，也实用户开机或者关机。一旦某个方格里面开机的移动电话数量发生了变化，该方格里的收发站就会向总部发送一条信息说明这个改变量。

总部要你写一个程序，用来管理从各个收发站收到的信息。老板可能随时会问：某个给定矩形区域内有多少部开机的移动电话啊？你的程序必需要能随时回答老板的问题。

【输入输出数据】

从标准输入读入整数，向标准输出写入你对老板的回答。

输入数据的格式例如以下：每一个输入独立成一行。一个输入包含一个指示数和一些參数，见下表：

指示数	參数	意义
0	S	初始指令。整个区域由S*S个小方格组成。这个指令仅仅会在一開始出现一次。
1	X Y A	方格（X，Y）内的开机移动电话量添加了A。A可能是正数也可能是负数
2	L B R T	询问在矩形区域(L,B)—(R,T)内有多少部开机的移动电话。矩形区域(L,B)—(R,T)包含全部的格子（X。Y）满足L<=X<=R,B<=Y<=T.
3		终止程序。这个指令仅仅会在最后出现一次。

全部的数据总是在给定范围内，你不须要差错。

特别的，假设A是负数，你能够觉得该操作不会让该格子的开机移动电话数变成负数。格子是从0開始编号的，比方一个4*4的区域，全部的格子（X，Y）应该表示为：0<=X<=3,0<=Y<=3。

对于除了2之外的指示，你的程序不应该输出不论什么东西。

假设指示是2。那么你的程序应该向标准输出写入一个整数。

【数据限制】

区域大小	S * S	1 * 1 <= S * S <= 1024 * 1024
每一个格子的值	V	0 <= V <= 32767
添加/降低量	A	-32768 <= A <= 32767
指令总数	U	3 <= U <= 60002
全部格子的和	M	M= 2^30

第一次在OJ题中用函数指针数组。感觉挺简洁的。

#include <stdio.h>

int tree[1030][1030], size;

int lowBit(int n)
{
	return n & (-n);
}

void getSize()
{
	scanf("%d", &size);
}

void update()
{
	int x, y, val;
	scanf("%d%d%d", &x, &y, &val);
	++x; ++y;
	
	int temp;
	while(x <= size){
		temp = y;
		while(temp <= size){
			tree[x][temp] += val;
			temp += lowBit(temp);
		}
		x += lowBit(x);
	}
}

int getSum(int x, int y)
{
	int sum = 0, temp;
	while(x > 0){
		temp = y;
		while(temp > 0){
			sum += tree[x][temp];
			temp -= lowBit(temp);
		}
		x -= lowBit(x);
	}
	return sum;
}

void query()
{
	int x1, y1, x2, y2;
	scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
	++x1; ++y1; ++x2; ++y2;
	
	int sum = getSum(x2, y2) - getSum(x2, y1 - 1) - 
		getSum(x1 - 1, y2) + getSum(x1 - 1, y1 - 1);
	printf("%d
", sum);
}

void (*funArr[])() = {
	getSize, update, query
}; //函数指针数组

int main()
{
	int com;
	while(scanf("%d", &com), com != 3)
		(*funArr[com])();	
	return 0;
}