Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14489   Accepted: 6735 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14391   Accepted: 6685 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 13774   Accepted: 6393 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

树状数组支持两种操作: Add(x, d)操作:   让a[x]增加d. Query(L,R): 计算 a[L]+a[L+1]……a[R]. 当要频繁的对数组元素进行修改,同时又要频繁的查询数组内任一区间元素之和的时候,可以考虑使用树状数组. 通常对一维数组最直接的算法可以在O(1)时间内完成一次修改,但是需要O(n)时间来进行一次查询.而树状数组的修改和查询均可在O(log(n))的时间内完成. 在二维情况下:数组A[][]的树状数组定义为: C[x][y] = ∑ a[i][j], 其中, …

题链: http://poj.org/problem?id=1195 题解: 二维树状数组 #include<cstdio> #include<cstring> #include<iostream> #define MAXN 1500 using namespace std; struct BIT{ int val[MAXN][MAXN],N; void Reset(int n){N=n; memset(val,0,sizeof(val));} int Lowbit(i…

题目链接:http://poj.org/problem?id=1195 题目意思:有一部 mobie phone 基站,它的面积被分成一个个小正方形(1 * 1 的大小),所有的小正方形的面积构成了一个 S * S 大小的矩阵(下标都是从 0 ~ S-1 变化的). 有四种指令: 第 一 行的指令默认输入是 0, 空格之后是矩阵的大小: S 最后一行的指令是 3, 代表 整个输入结束 注意:这两行的指令只会出现一次! 夹在它们中间的指令有可能是指令1,假设为X Y A,代表向第 X 行 第 Y…

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 contain…

二维的树状数组,,, 记得矩阵的求和运算要想好在写.... 代码如下: #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <cstdlib> #include <stack> #include <queue> #include <vector> #include <algorithm>…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 18489   Accepted: 8558 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

树状数组,开始的时候wa了,后来看看,原来是概率论没学好,以为求(L,B) - (R,T) 矩阵内的和只要用sum(R+1,T+1) - sum(L,B) 就行了,.傻x了.. 必须 sum(R,T) - sum(L,T) - sum(R,B) + sum(L,B) ;  (R,T 已经自加1)   诫之. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…