x1 vs directv

OpenCV -- Skewing // define head function#ifndef PS_ALGORITHM_H_INCLUDED#define PS_ALGORITHM_H_INCLUDED#include #include #include "cv.h"#include "highgui.h"#include "cxmat.hpp"#include "cxcore.hpp"#include "math.h"using namespace std;using namespace cv;void Show_Image(Mat, const string );#endif // PS_ALGORITHM_H_INCLUDED#include "PS_Algorithm.h"#include using namespace std;using namespace cv;#define pi 3.1415926int main(){ string Img_name("4.jpg"); Mat Img; Img=imre

There are many online draw algorithms, but it is not easy to achieve high speed and simplicity. Slope-phase multiplication is one of the simplest methods, but it takes a lot of time to calculate each point for multiplication and division operations. The following describes Bresenham's efficient line-drawing algorithm, you only need to add or subtract the coordinates of each vertex.The simplified algorithm is described in the pseudo Pascal language as follows:Procedure DrawLine (

{ scanf("%s", b); sscanf(b, "(%lf,%lf)", x[N][i], y[N][i]); } x[N][3] = x[N][0], y[N][3] = y[N][0]; }else { scanf("%d", vn[N]);for(i = 0; i { scanf("%s", b); sscanf(b, "(%lf,%lf)", x[N][i], y[N][i]); } x[N][vn[N]] = x[N][0], y[N][vn[N]] = y[N][0]; }}int init(){int i, j, k;for(N = 0; ; N ++) { scanf("%s", b);if(b[0] == '.')return 0;if(b[0] == '-')break; input(); }return 1;}

Point and Rectangle collision/** * * @param x1 Point * @param y1 point * @param x2 rectangle View x * @param y2 rectangle View y * @param w Rectangle View Width * @param h Rectangle v Iew High * @return */public static boolean iscollsion (int x1, int y1, int x2, int y2, int w, int h) {if (X1 >= x2 am P X1 Rectangula

-1) {THIS.MAPTITLE[I][J] = 1;} else {THIS.MAPTITLE[I][J] = 0;}}}This.width = (This.stage.stageWidth-(this.maptitle[0].length + 1) * This.strokewid)/this.maptitle[0].length;This.map = new Mapclass (this.maptitle);This.initobstacle ();For (Let I in This.mapobstacle) {This.map.update (This.mapobstacle[i][0], this.mapobstacle[i][1], True)}}/*** Show obstacles on the map* */Protected initobstacle (): void {for (Let i = 0; i if (This.mapobstacle[i][0] This.mapobstacle[i][1] THIS.MAPTITLE[THIS.MAPOBST

pi+1 6 on the visible side of the window edge . Else computes the intersection point and outputs intersection 7). else if pi+1 on the visible side of the window, then compute the intersection point and output the intersection point, outputting pi+1 end 8). End of Algorithm Four Key code: 1. Encode the region: int Ccg_wuping2dtransview::p code (int *x, int *y) { int code = 0; ccg_wuping2dtransdoc* PDoc = GetDocument (); if (*x 2.cohen-sutherland Line cropping: int ccg

is no displacement before and after the transformation), and the coefficient i is adjusted to 1, so there is where p is the scale factor Write an equation PX ' = Ax + by (1) Py ' = Dx + Ey (2) p = Gx + Hy + 1 (3) In the Formula 3 elimination 1 and the formula 2 p, you get 2 equations Ax + by-gxx '-hyx ' = X ' (4) Dx + ey-gxy '-Hyy ' = Y ' (5) We assume that the coordinates of the unit square area consisting of point (0,0) (1,0) (0,1) (four) points are mapped to the target plane, respectively