POJ 1195 Mobile phones

Mobile phones

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 11530 Accepted: 5284

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 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

34

Source

IOI 2001

解题思路：不能直接求和，否则会超时。用树状数组。

  123

4

  0123100000210230320-100430000

以第二次操作为例：表中最外围行列是行列的值，向内一行（列）是数组下标，构成一个4×4的二维数组，进行树状数组操作时按外围行列计算。

Code:

#include<iostream>using namespace std;int S[1200][1200];int s;int lowbit(int x)//将x转化为2进制，其2进制下末尾0个数记为k，返回2^k{return x&(-x);}void Add(int p,int q,int aim)//树状数组加操作{for(int i=p;i<=s;i+=lowbit(i))for(int j=q;j<=s;j+=lowbit(j))            S[i][j]+=aim;}long long Sum(int p,int q)//树状数组求和操作{long long ans=0;for(int i=p;i>0;i-=lowbit(i))for(int j=q;j>0;j-=lowbit(j))ans+=S[i][j];return ans;}long long Ans(int x1,int y1,int x2,int y2){return Sum(x1,y1)+Sum(x2+1,y2+1)-Sum(x1,y2+1)-Sum(x2+1,y1);}int main(){int a,b,i,j,l,r,t,x,y,operation;long long M;while(cin>>operation&&operation!=3){switch(operation){case 0:cin>>s;  for(i=0;i<=s;i++)  for(j=0;j<=s;j++)  S[i][j]=0;  break;case 1:cin>>x>>y>>a;  Add(x+1,y+1,a);  break;case 2:cin>>l>>b>>r>>t;M=Ans(l,b,r,t);cout<<M<<endl;break;}}return 0;}

有关树状数组，请看http://download.csdn.net/detail/hqu_fritz/4708969和http://download.csdn.net/detail/hqu_fritz/4695317