Mobile phones(二维树状数组)

传送门:Poj-1195

题目描述

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.

输入

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:$M=2^{30}$

样例

  • 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
  • Output
    3
    4

题解

  • 题意: 1——在(x,y)加上A ; 2——求左上角为(L,B),右下角为(R,T)的矩阵和
  • 裸的二维树状数组。二维数组的思路和一维类似,一维中第x个元素的值为x的前缀和,而二维中(x,y)的值为以(0,0)为左上角,以(x,y)为右下角的矩阵和(这里的求和都是以树状数组lowbit规则计算的)。

Code

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//https://vjudge.net/contest/244678#problem/D
#include<stdio.h>
#include<string.h>
#define LL long long int
#define INIT(a,b) memset(a,b,sizeof(a))
const int MAX=0x7fffffff;
const int MIN=-0x7fffffff;
const int INF=0x3f3f3f3f;
using namespace std;
int tree[1030][1030];
int que,s,x,y,a,l,b,r,t;
int lowbit(int x){return x & -x;}
void update(int n,int m,int t){
for(int i=n;i<=s;i+=lowbit(i))
for(int j=m;j<=s;j+=lowbit(j))
tree[i][j]+=t;
}
int sum(int n,int m){
int cnt=0;
for(int i=n;i>0;i-=lowbit(i))
for(int j=m;j>0;j-=lowbit(j))
cnt+=tree[i][j];
return cnt;
}
int main()
{
while(~scanf("%d",&que)){
if(que==0){
scanf("%d",&s);
INIT(tree,0);
}
else if(que==1){
scanf("%d %d %d",&x,&y,&a);
x++;y++;
update(x,y,a);
}
else if(que==2){
scanf("%d %d %d %d",&l,&b,&r,&t);
printf("%d\n",sum(r+1,t+1)-sum(r+1,b)-sum(l,t+1)+sum(l,b));
}
else return 0;
}
return 0;
}
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