?id=1195
Mobile phones
Time Limit:5000MSMemory Limit:65536K
Total Submissions:16721Accepted:7692
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 * 1024Cell value V at any time: 0 <= V <= 32767Update amount: -32768 <= A <= 32767No of instructions in input: 3 <= U <= 60002Maximum 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
思路: 二维树状数组
分析:1 给定一个矩阵和三种操作 1 a b x表示把[a,b]值加上x，2 L B R T表示L <= x <= R , B <= y <= T求这个小矩形的面积 3表示输入结束2 简单的二维树状数组，，但是这边要注意的是操作2的时候一定不能够把维度给弄反了，L R表示的是x的范围而B T表示的是y的范围3 题目还有一个trick就是当[a,b]加上x后如果值为负数，那么应该把值设为0
代码:
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 1030;int s;int num[MAXN][MAXN];long long treeNum[MAXN][MAXN];int lowbit(int x){return x&(-x);}void add(int x , int y , int val){if(num[x][y]+val < 0)val = -num[x][y];num[x][y] += val;for(int i = x ; i < MAXN ; i += lowbit(i))for(int j = y ; j < MAXN ; j += lowbit(j))treeNum[i][j] += val; }long long getSum(int x , int y){long long sum = 0;for(int i = x ; i > 0 ; i -= lowbit(i))for(int j = y ; j > 0 ; j -= lowbit(j))sum += treeNum[i][j];return sum;}int main(){int x , i , j , val;int l , r , u , d;while(scanf("%d%d" , &x , &s) != EOF){memset(treeNum , 0 , sizeof(treeNum));memset(num , 0 , sizeof(num));while(scanf("%d" , &x) && x != 3){if(x == 1){scanf("%d%d%d" , &i , &j , &val);add(i+1 , j+1 , val);}else{scanf("%d%d%d%d" , &u , &l , &d , &r);l++ , u++ , r++ , d++;long long sum = getSum(d , r);sum -= getSum(u-1 , r);sum -= getSum(d , l-1);sum += getSum(u-1 , l-1);printf("%lld\n" , sum);}}}return 0;}
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