matrices precalculus

Symmetric matrix: An element that corresponds to an equal matrix of the main diagonal axisThe symmetric matrix storage uses the upper triangle, the lower triangle storage, the lower triangle conforms to the i>=j (row is greater than equals the column), I use the lower triangle storage, the storage may store the lower triangle the element, but also needs to turn it to the symmetry matrix form printing when printingconstructor function:Symmetricmatrix (t* a,size_t size): _a (New t[size* (size+1)/2

Sparse matrix: M*n Matrix, the number of valid values in the matrix is much smaller than the number of invalid values, and these number distribution is not regular.Compressed storage has very few values and uses triples (VALUE,ROW,COL) to store each valid value. The ternary group is stored in line priority according to the position of the original matrix.constructor function:Sparsematrix (t* a,size_t m,size_t n,const t Invalid): _rowsize (M), _colsize (n), _invalid (invalid) {for (size_t i=0;i P

#include This article is from the "sunshine225" blog, make sure to keep this source http://10707460.blog.51cto.com/10697460/1758809Storage and fast transpose of matrices

Shear is the transformation of the coordinate system, non-uniform stretching. When shearing, the angle changes, but the area or volume does not change. It can also be understood as the angle change between axes, resulting in distortions.For example, this is the x-coordinate of the y-coordinate of the shear, the robot's Y coordinate does not change, only the x-coordinate changes, the changed coordinates x ' can be understood as the y-coordinate multiplied by the shear factor s and the original co

Analysis we have solved one-dimensional problems (the largest sub-segments and problems in the foundation), and now we are two-dimensional, and we see if we can turn this problem into one-dimensional problem. The last sub-matrix must be between some two rows. Suppose we think that the sub-matrix is between line I and column J, how do we get I and J, yes, enum.Enumerates all 1For i = 1 to m dofor j = i to M do//calculates the first row I to column j and for k = 1 to N doc[k] = (j = = i)? A[i][k]:

You is given the number of rows and number of n_rows columns of n_cols a 2D binary matrix where all values is 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, col.id] of that value.Also, write a function reset which sets all values 0. Try to minimize the number of calls to System ' s math.random () and optimize the time and space complexity.Note: 1 0 and0 flipWon't is called when the matrix h

; } } returnmaxsum;}intMain () {inta[8] = {4,-3,-4,8,-5,7,-5,7}; inti,i1,j1; printf ("%d-->%d:%d\n", I1,j1,sub_sum (A,8, i1,j1)); return 0;}Maximum sub-matrices:Problem Description: Given a n*m matrix A, a non-empty matrix in a, so that the element and maximum in this sub-matrix. Wherein, A's sub-matrix refers to a row and column in a continuous block.Idea: fixed the range of column I to section J, looking for the maximal sub-matrix in this range, that is, the elements on column I of each

The adjacent knot matrix is simple, its bfs,dfs, two kinds of traversal is also relatively simple, one with the queue, one with the array can!!!But the adjacency matrix is a waste of space, especially when it is a sparse matrix!!!--------------------------------------------------------------------------------------------------------------- ------------------------Establishment of adjacency matrix and its BFS, DFS, traversal # include Establishment and Bfs,dfs of adjacent knot

When a matrix is a sparse matrix, the number of valid data is much less than the invalid data, so if a matrix all storage will waste space, you can only store valid data, invalid data as a markerThe code is as follows:#include This article is from the "Knock Code good Sleep zzz" blog, please be sure to keep this source http://2627lounuo.blog.51cto.com/10696599/1750630Compact storage and fast transpose of sparse matrices

I want to use simple template matching for image recognition.The pre-processed character picture is matched to a sample picture of A to J, respectively. The most significant results indicate that the character picture can be identified.In practice, I used the OpenCV matchtemplate () function, but I didn't get the effect I wanted.The function of Matchtemplate () is to search for the specified template in the image, if the template is intercepted from the image to be searched, it will have a good

#include "stdafx.h"#include #include #include void Inputscore (int *p,int m,int n);int Findmax (int *p,int m,int n,int *prow,int *pcol);int main (){int *pscore,m,n,maxscore,row,col;printf_s ("Input array size m,n:");scanf_s ("%d,%d", m,n,sizeof (m));Pscore = (int *) calloc (m*n,sizeof (int));if (Pscore = = NULL){printf_s ("No enough memory!\n");Exit (0);}Inputscore (Pscore,m,n);Maxscore = Findmax (Pscore,m,n,row,col);printf_s ("Maxscore =%d,class =%d,number =%d\n", maxscore,row+1,col+1);Free (Ps

Just started messing around. Network flow solver, assuming that the maximum flow = all elements of the and then there is a solution, the use of residual network inference is unique,//method There are two, the first is to deep search to see if there is a positive edge of the ring. See the previous article 4888//at least four points form the ring, the other is to use a matrix DP, only need to satisfy a row of the I-column element Copyright notice: This article blog original article. Blogs, withou

string.h>4 using namespacestd;5 inta[ the][ the],b[ the];6 intn,cursum=- the,max=numeric_limitsint>:: Min ();7 8 intcurmaxsum ()9 {Ten intsum=0; One for(intI=0; i) A { -sum+=B[i]; - if(sum0) thesum=B[i]; - if(sum>cursum) -cursum=sum; - } + returncursum; - } + A intmaxsub () at { - for(intI=0; i) - { -memset (b,0,sizeof(b)); - for(intJ=i; j) - { in //printf ("\nj=%d\n", j); - for(intk=0; k) to {

. It can be found that each column of V is Eigenvectors , each of the columns of U is feature vectors, each of which is emitted in descending order of the characteristic values. What problems can SVD solve? The first is to solve the PCA, you can findThe matrix to the left of this equation is actually an x this data set (already preprocessed, the mean is 0) the covariance matrix (set to M) of the first k eigenvalues corresponding to the characteristic vector transpose is arranged. =m*m (M is the

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties: Integers in each row is sorted from the left to the right. The first integer of each row was greater than the last integer of the previous row. For example,Consider the following matrix:[ [1, 3, 5, 7], [Ten, One,], [23, 30, 34, 50]]Given Target = 3 , return true .Idea: This problem is still relatively simple, mainly with two dichotomy, only for the first column

Using adjacency matrix to create a forward net NAnalysis: The creation of the graph is mainly using the input vertices, and stored in a vector (one-dimensional array), and then by entering two vertices and weights to create arcs, using a two-dimensional array representation, therefore, the use of adjacency matrix to create a graph storage requires two arrays: a one-dimensional array and a two-dimensional array.#include Copyright NOTICE: This article for Bo Master original article, without Bo Ma

(t--) {scanf ("%d",N); intm=0; for(i=0; i) {scanf ("%LF%LF%LF%LF",x1,y1,x2,y2); S[M].L=x1;s[m].r=x2;s[m].h=y1;s[m].f=1; mark[m++]=X1; S[M].L=x1;s[m].r=x2;s[m].h=y2;s[m].f=-1; mark[m++]=x2; } sort (S,s+m,cmp); Sort (Mark,mark+m); intk=1; for(i=1; i) { if(mark[i]!=mark[i-1]) Mark[k++]=Mark[i]; } /*for (i=0;i*/Build (0, K-1,1); Doubleans=0; for(i=0; i) { intLl=find (S[I].L,0, K-1); intRr=find (S[I].R,0, K-1)-1; Updata (LL,RR,S[I].F,0, K-1,1); Ans+ = (s[

Programming Question # #: Calculating the sum of the edge elements of matricesSource: POJ (Coursera statement: The exercises completed on POJ will not count against the final results of Coursera. )Note: Total time limit: 1000ms memory limit: 65536kBDescribeEnter an integer matrix to calculate the sum of the elements at the edge of the matrix. The elements of the so-called Matrix Edge are the elements of the first and last lines, and the elements of the first and last columns.InputThe first behav

#include "stdafx.h" #include 　　Inversion of sparse matrices (normal inversion and normal inversion)

distribution, that is, the mean value is 0 and the standard deviation is 1.three. L2 StandardizationThus, we can write the simplest MATLAB source code as follows: first, by line normalization:% examplesa=[3 4;5 12]; [m n] = size (A);% normalize each row to unitfor i = 1:m A (i,:) =a (i,:)/norm (A (I,:)); endNormalized by column.% normalize each column to unita=[3 4;5 12];for i = 1:n A (:, i) =a (:, i)/norm (A (:, i)); endHowever, the code above is best for functionality, but not optimal,