c++ - 抗锯齿 PPM 文件

Question

我从下面的代码中得到了这个输出 circle.ppm它有很多锯齿状的边缘。我相信我需要某种类型的抗锯齿方法，但我很难开始。

摆脱它们的最简单方法是什么？

//******************** INCLUDE: ********************

#include <fstream>                      // ofstream
#include <iostream>                     // cin, cout
#include <cmath>                        // sqrt, pow

using namespace std;                    // USING STANDARD NAMESPACE

//******************** DECLARATION OF FUNCTIONS: ********************

bool isInsideCircle(int x,int y);       // CHECKS IF (X,Y) IS INSIDE CIRCLE
int Brightness(int x,int y);            // DISTANCE FROM CENTER


//******************** VARIABLES: ********************

const int WIDTH = 255, HEIGHT = 255;    // WIDTH AND HEIGHT RESOLUTION
const int RADIUS = 100;                 // RADIUS OF CIRCLE

const int CENTER_X = 0.5 * WIDTH;       // CENTER X-COORDINATES
const int CENTER_Y = 0.5 * HEIGHT;      // CENTER Y-COORDINATES

int r, g, b;                            // RED, GREEN AND BLUE

//******************** MAIN FUNCTION: ********************

int main() {

  system("open circle.ppm");            // OPENS THE IMAGE
  system("killall ToyViewer");          // CLOSES THE IMAGE

//******************** PPM SETUP: ********************

  ofstream img ("circle.ppm");          // MAKE THE FILE
  img << "P3" << endl;                  // MAGIC NUMBER
  img << WIDTH << endl;                 // WIDTH RESOLUTION
  img << HEIGHT << endl;                // HEIGHT RESOLUTION
  img << "255" << endl;                 // 0-255 MAX RGB VALUE

//******************** FOR EACH PIXEL: ********************

  for (int y = 0; y < HEIGHT; y++) {

      for (int x = 0; x < WIDTH; x++) {

          if (isInsideCircle(x, y)) {

              r = 50 + Brightness(x, y) % 255;
              g = 50 + Brightness(x, y) % 255;
              b = 50 + Brightness(x, y) % 255;
          }

          else {

              r = 0;
              g = 0;
              b = 0;

          }

          img << r << " " << g << " " << b << endl;
      }
  }

//******************** PROGRAM END: ********************

  system("open circle.ppm"); 
  return 0;

}

//******************** FUNCTIONS: ********************

bool isInsideCircle(int x, int y) {

  if (pow((x - CENTER_X),2) + pow((y - CENTER_Y),2) < pow(RADIUS,2)) {

      return true;

  } else {

      return false;
  }
}

int Brightness(int x, int y) {

  return abs( sqrt( pow((x - CENTER_X),2) + pow((y - CENTER_Y),2)) - RADIUS);

}

Answer 1

正如 Daniel 指出的那样，您的收入急剧下降。实际上不是从 1 到 0，而是从 {50,50,50} 到 {0,0,0}。那是因为 50 + Brightness 行。

简单的解决方案是使用稍微不那么急剧的下降。定义一个从 50 渐近下降到 0 的 Brightness2(x,y)。

int Brightness2(int x, int y) {
  int dx = x - CENTER_X;
  int dy = x - CENTER_Y;
  return 50/(1+(dx*dx+dy*dy-RADIUS*RADIUS));
}