0 引言 问题背景:给定CAD模型与点云的情况下,基于CAD比对实现对实测点云的尺寸测量是三维测量技术中一种常用的思路.该思路的关键问题在于需要精确求解CAD模型与点云之间的空间转换矩阵.采用ICP(迭代最近邻点算法)的变种,将基于点-点对应关系的算法拓展为基于平面-平面的算法.当前,基于二次开发技术求解出了CAD模型中的所有平面,并且基于点云PCL库中的生长聚类算法(regionGrowing做分割,RANSAC对平面方程进行拟合)提取出了典型平面,因此希望通过平面-平面对应关系解出二者之间的

Spiral Matrix II 螺旋矩阵 Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. For example,Given n = 3, You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ] 看博客园有人面试碰到过这个问题,长篇大论看得我头晕

Today we have learned the Matrix Factorization, and I want to record my study notes. Some kownledge which I have learned before is forgot...(呜呜) 1.Terminology 单位矩阵:identity matrix 特征值:eigenvalues 特征向量:eigenvectors 矩阵的秩:rank 对角矩阵:diagonal matrix 对角化矩阵

http://acm.hdu.edu.cn/showproblem.php?pid=4965 2014 Multi-University Training Contest 9 1006 Fast Matrix Calculation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 238    Accepted Submission(

设S[k] = A + A^2 +````+A^k. 设矩阵T = A[1] 0 E E 这里的E为n*n单位方阵,0为n*n方阵 令A[k] = A ^ k 矩阵B[k] = A[k+1] S[k] 则有递推式B[K] = T*B[k-1],即有B[k] = T^k*B[0],令S[0] 为n*n的0矩阵. 矩阵快速幂求出即可····· 还可以使用两次分治的方法····自行百度···· 贴代码: #include<cstdio> #include<cstring> int n,k

题目: 矩阵的之字型遍历 给你一个包含 m x n 个元素的矩阵 (m 行, n 列), 求该矩阵的之字型遍历. 样例 对于如下矩阵: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10, 11, 12] ] 返回 [1, 2, 5, 9, 6, 3, 4, 7, 10, 11, 8, 12] 解题: 感觉很简单,就是没有搞出来,程序来源 ,这里是先右上走,我换成先横着走就是不对了,表示不理解.另外一种方法,表示也不理解. java程序: public class Solut

意甲冠军  由于矩阵乘法计算链表达的数量,需要的计算  后的电流等于行的矩阵的矩阵的列数  他们乘足够的人才  非法输出error 输入是严格合法的  即使仅仅有两个相乘也会用括号括起来  并且括号中最多有两个 那么就非常easy了 遇到字母直接入栈  遇到反括号计算后入栈  然后就得到结果了 #include<cstdio> #include<cctype> #include<cstring> using namespace std; const int N = 10

Matrix multiplication                                                                           Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given two matrices A and B of size n×n, find the

职务地址:POJ 3233 题目大意:给定矩阵A,求A + A^2 + A^3 + - + A^k的结果(两个矩阵相加就是相应位置分别相加).输出的数据mod m. k<=10^9.     这道题两次二分,相当经典.首先我们知道,A^i能够二分求出. 然后我们须要对整个题目的数据规模k进行二分.比方,当k=6时,有:     A + A^2 + A^3 + A^4 + A^5 + A^6 =(A + A^2 + A^3) + A^3*(A + A^2 + A^3)     应用这个式子后,规模

给定一个矩阵,把零值所在的行和列都置为零.例如: 1 2 3 1 3 1 1 1 操作之后变为 1 3 0 0 0 1 1 方法1: 赋值另存一个m*n的矩阵,在原矩阵为零的值相应置新的矩阵行和列为零.额外空间为O(m*n). 方法2: 两个数组,bool[m] 和 bool[n] 分别存某行有零,后者某列有零.之后根据数组值将原矩阵相应位置置零.额外空间O(m + n). class Solution { public: void setZeroes(vector<vector<int>

题目背景 SOURCE:NOIP2016-RZZ-1 题目描述 给出两个 N×N 的矩阵 A.B,矩阵每行每列标号 0-N-1 .定义这两个矩阵的乘积 AB 为

In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data. You're given a matrix represented by a two-dimensional array, and two positive integers r and c rep

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. click to show follow up. Follow up: Did you use extra space?A straight forward solution using O(mn) space is probably a bad idea.A simple improvement uses O

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example,Given the following matrix: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] You should return [1,2,3,6,9,8,7,4,5]. 题目标签:Array 这道题目给了我们一个矩阵,

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. The distance between two adjacent cells is 1. Example 1: Input: 0 0 0 0 1 0 0 0 0 Output: 0 0 0 0 1 0 0 0 0 Example 2: Input: 0 0 0 0 1 0 1 1 1 Output: 0 0 0 0 1 0

You are given the number of rows n_rows and number of columns n_cols of a 2D binary matrix where all values are initially 0. Write a function flip which chooses a 0 value uniformly at random, changes it to 1, and then returns the position [row.id, co

Matrix Power Series r时间限制: 1 Sec 内存限制: 512 MB 题目描述 给定矩阵A,求矩阵S=A^1+A^2+--+A^k,输出矩阵,S矩阵中每个元都要模m. 数据范围: n (n ≤ 30) , k (k ≤ 109) ,m (m < 104) 输入 输入三个正整数n,k,m 输出 输出矩阵S mod m 样例输入 2 2 4 0 1 1 1 样例输出 1 2 2 3 这道题不多说,可以得出加速矩阵(E为单位矩阵,也就是形为\(\begin{bmatrix}1&

233 Matrix Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1190    Accepted Submission(s): 700 Problem Description In our daily life we often use 233 to express our feelings. Actually, we may s

http://poj.org/problem?id=3233 题目大意:给定矩阵A,求A + A^2 + A^3 + … + A^k的结果(两个矩阵相加就是对应位置分别相加).输出的数据mod m.k<=10^9.这道题两次二分,相当经典.首先我们知道,A^i可以二分求出.然后我们需要对整个题目的数据规模k进行二分.比如,当k=6时,有:A + A^2 + A^3 + A^4 + A^5 + A^6 =(A + A^2 + A^3) + A^3*(A + A^2 + A^3)应用这个式子后,规模

1105 Spiral Matrix(25 分) This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The

题目:矩阵置0 难度:Easy 题目内容: Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place. 翻译: 给定一个m x n矩阵,如果一个元素是0,就把它的整行和列设为0. 要求:就地置0. Example 1: Input: [   [1,1,1],   [1,0,1],   [1,1,1] ] Output: [   [1,0,1],   [0,0,0],