C++實(shí)現(xiàn)基于EASYX庫掃描線算法
更新時間:2020年02月20日 12:51:30 作者:py粉
這篇文章主要為大家詳細(xì)介紹了C++實(shí)現(xiàn)基于EASYX庫掃描線算法,文中示例代碼介紹的非常詳細(xì),具有一定的參考價值,感興趣的小伙伴們可以參考一下
本文實(shí)例為大家分享了C++實(shí)現(xiàn)基于EASYX庫掃描線算法的具體代碼,供大家參考,具體內(nèi)容如下
掃描線算法的基本原理

* 作者在掃描線算法的基礎(chǔ)上自己設(shè)計的更易于理解的地物填充繪制算法
流程圖

代碼
#include<graphics.h>
//#include<conio.h>
#include<iostream>
using namespace std;
//-----------------------------草圖形-----------------------------//
void Grass(double x, double y, double hight, double width)
{
setlinecolor(GREEN);
line(x, y, x - hight / 2, y - width);
line(x, y, x, y - width);
line(x, y, x + hight / 2, y - width);
}
//-----------------------------林圖形-----------------------------//
void Tree(double x, double y, double hight, double width)
{
setlinecolor(GREEN);
line(x, y, x - hight / 2, y - width);
line(x, y, x + hight / 2, y - width);
line(x, y - width, x, y + width);
}
//------------------求解方程的k,b值---------------------//
void function_line(double x1, double y1, double x2, double y2, double &k, double &b)
{
if (x2 - x1 == 0) {
k = -9999;
b = x1;
return;
}
k = (y2 - y1) / (x2 - x1);
b = y1 - k * x1;
}
//------------------返回直線的y----------------------//
double solve_y(double k, double b, double x)
{
return k * x + b;
}
//------------------返回直線的x----------------------//
double solve_x(double k, double b, double y)
{
if (k == -9999)
return b;
return (y - b) / k;
}
//-------------------找出掃描線與邊界相交的點(diǎn),并儲存在suitable_x[]中,把相交的點(diǎn)數(shù)儲存在n中--------------//
//x[]-----------頂點(diǎn)x坐標(biāo)
//y[]-----------頂點(diǎn)y坐標(biāo)
//N-------------頂點(diǎn)的個數(shù)
//now_y---------當(dāng)前的掃描線y
//suitable_x[]--掃描線與邊界相交的點(diǎn)的x坐標(biāo)
//n-------------相交的點(diǎn)數(shù)
void Suitable_x(double x[], double y[],int N, double now_y, double suitable_x[], int &n)
{
n = 0;
int t = 0;
double k, b;
for (int i = 0; i < N; i++)
{
if (i == N - 1)
{
if (y[i] < now_y&& now_y <= y[0] || y[0] < now_y&& now_y <= y[i])//判斷直線是否過最后一條直線
{
function_line(x[0], y[0], x[i], y[i], k, b);
suitable_x[t++] = solve_x(k, b, now_y);
}
}
else
{
if (y[i] < now_y&& now_y <= y[i + 1] || y[i + 1] < now_y&& now_y <= y[i]) //判斷掃描線過哪條直線的范圍
{
function_line(x[i], y[i], x[i + 1], y[i + 1], k, b);
suitable_x[t++] = solve_x(k, b, now_y);//把交點(diǎn)的x儲存在suitable_x[]中
}
}
}
n = t;
}
void sort_min_to_max(double suitable_x[],int n) //排序----從小到大排序
{
double temp;
for (int i = 0; i < n - 1; i++)
{
for (int j = i + 1; j < n; j++) {
if (suitable_x[i] > suitable_x[j]) {
double temp = suitable_x[i];
suitable_x[i] = suitable_x[j];
suitable_x[j] = temp;
}
}
}
}
double find_max(double elem[],int N) //輸出最大的元素值
{
double max = elem[0];
for (int i = 0; i < N; i++)
{
if (elem[i] > max)
{
max = elem[i];
}
}
return max;
}
double find_min(double elem[],int N) //輸出最小的元素值
{
double min = elem[0];
for (int i = 0; i < N; i++)
{
if (elem[i] < min)
{
min = elem[i];
}
}
return min;
}
int main()
{
cout << "-----------------------------------------地物填充繪制算法--------------------------------------------\n\n";
cout << " 創(chuàng)建人:李景勃\n";
cout << " 時間:2018年10月24日\n";
cout << " 10月24日是全部程序員值得開心的節(jié)日,也是自己重要的日子,23號完成,祝福全部程序員們?。?! \n\n\n";
int N,ch;
char color;
double x[50],y[50];
cout << "輸入填充的多邊形的頂點(diǎn)數(shù):";
cin >> N;
cout << "依次輸入地物的頂點(diǎn)坐標(biāo):";
for (int i = 0; i < N; i++)
{
cout << "輸入第" << i+1 << "個頂點(diǎn)的坐標(biāo):" << endl;
cout << "x,y:";
scanf_s("%lf,%lf",&x[i],&y[i]);
/*cout << "x = "; cin >> x[i];
cout << "y = "; cin >> y[i];*/
}
cout << "選擇您填充的地物: 整個區(qū)域單顏色充填---請按0;草地---請按 1;林地---請按 2 ; "; cin >> ch;
if (ch == 0)
{
cout << "請選擇顏色:BLACK 黑,DARKGRAY 深灰綠色 BLUE 藍(lán) LIGHTBLUE 亮藍(lán):GREEN 綠 CYAN 青 RED 紅 MAGENTA 紫 BROWN 棕 YELLOW 黃 LIGHTGRAY 淺灰---注意是大寫的英文\n ";
cout << "如果想讓顏色變亮請在顏色的前面加上LIGHT \n";
cout << "請輸入顏色:";
cin >> color;
}
initgraph(find_max(x, N)+100, find_max(y, N)+100);
//int N = 9;
//double x[50] = { 200,100,150,170,400,350,300,250,270 };
//double y[50] = { 200,300,400,250,350,300,250,100,270 };
double new_y[50] = { 0 };
int n = 0;
double suitable_x[10];
//草的參數(shù)
int hight = 12,
width = 6;
int space = 4; //草之間的間隔
setbkcolor(WHITE); //設(shè)置背景色
cleardevice(); //用背景色填充屏幕
setlinecolor(BLACK); //多邊形的線條顏色
for (int i = 0; i < N; i++) //逆時針頂點(diǎn)相連構(gòu)成閉合多邊形
{
if (i == N - 1)
{
line(x[i], y[i], x[0], y[0]);
}
else
{
line(x[i], y[i], x[i + 1], y[i + 1]);
}
}
double min_y = find_min(y,N),
max_y = find_max(y,N);
switch (ch)
{
case 0:
//-------------------------像素填充面積-------------------------//
for (int now_y = min_y; now_y <= max_y; now_y++)
{
Suitable_x(x, y, N, now_y, suitable_x, n);
sort_min_to_max(suitable_x, n); //從小到大排序
for (int i = 0; i < n; i += 2)
{
for (int now_x = suitable_x[i]; now_x < suitable_x[i + 1]; now_x++)//確定在掃描線上,要填充的范圍
{
putpixel(now_x, now_y, color);
}
}
}
break;
case 1:
//-------------------------草填充-------------------------//
for (int now_y = min_y + width + space; now_y <= max_y; now_y += width + space)
{
Suitable_x(x, y, N, now_y, suitable_x, n);
sort_min_to_max(suitable_x, n); //從小到大排序
for (int i = 0; i < n; i += 2)
{
for (int now_x = suitable_x[i] + hight / 2 + space + space; now_x < suitable_x[i + 1] - hight / 2; now_x += hight + space)//確定在掃描線上,要填充的范圍
{
Grass(now_x, now_y, hight, width);
Sleep(60);
}
}
}
break;
case 2:
//-------------------------樹填充-------------------------//
for (int now_y = min_y + width + space; now_y <= max_y; now_y += 2 * width + space)
{
Suitable_x(x, y, N, now_y, suitable_x, n);
sort_min_to_max(suitable_x, n); //從小到大排序
for (int i = 0; i < n; i += 2)
{
double longth = suitable_x[i + 1] - suitable_x[i]-space;
int num = longth / (hight+space);
//double space_tmp = longth - num * (hight + space);
//double space_ = space + space / num;
for (int now_x = suitable_x[i] + hight / 2 +space+space; now_x < suitable_x[i + 1] - hight / 2&&num--; now_x += hight + space)//確定在掃描線上,要填充的范圍
{
Tree(now_x, now_y, hight, width);
Sleep(60);
}
}
}
break;
default:
cout << "填充失敗?。?!" << endl;
break;
}
//closegraph();//關(guān)閉畫框
system("pause");
return 0;
}
實(shí)現(xiàn)效果




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