C++實現常用的平面計算幾何問題求解
來源:程序員人生 發布時間:2015-01-22 08:25:30 閱讀次數:2872次
通過封裝經常使用的點、線段類型,并提供點、線間的相互關系運算,為計算幾何工具庫的編寫提供基礎框架。
代碼以下:(代碼正確性仍需測試,謹慎使用)
//參考
//http://dev.gameres.com/Program/Abstract/Geometry.htm
//http://zhan.renren.com/jisuanjihe?from=template&checked=true
/*
toolbox: Geometry algorithm toolbox
author: alaclp
email: alaclp@qq.com
publish date: 2015⑴⑴6
*/
#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
//預定義
#define Min(x, y) ((x) < (y) ? (x) : (y))
#define Max(x, y) ((x) > (y) ? (x) : (y))
//點對象
typedef struct Point {
double x, y;
//構造函數
Point(double x, double y) : x(x), y(y) {}
//無參數時的構造函數
Point() : x(0), y(0) {}
//獲得到點pt的距離
double distance(const Point& pt) {
return sqrt( (x - pt.x) * (x - pt.x) + (y - pt.y) * (y - pt.y));
}
//判斷兩點是不是同1個點
bool equal(const Point& pt) {
return ((x - pt.x) == 0) && (y - pt.y == 0);
}
} Point;
//線段對象
typedef struct PartLine {
Point pa, pb;
double length;
PartLine() {
length = 0;
}
//構造函數
PartLine(Point pa, Point pb) : pa(pa), pb(pb) {
length = sqrt((pa.x - pb.x) * (pa.x - pb.x) + (pa.y - pb.y) * (pa.y - pb.y));
}
void assign(const PartLine& pl) {
pa = pl.pa;
pb = pl.pb;
length = pl.length;
}
//利用叉積計算點到線段的垂直距離
//注意:此結果距離有正負之分
//若pc點在線段的逆時針方向,則距離為正;否則,距離為副值
double getDistantToPoint(Point pc) {
double area = crossProd(pc) / 2;
return area * 2 / length;
/* 利用海倫公式計算
PartLine pl1(this->pa, pc), pl2(this->pb, pc);
double l1 = this->length, l2 = pl1.length, l3 = pl2.length;
double s = (l1 + l2 + l3) / 2; //海倫公式
double area = sqrt(s * (s - l1) * (s - l2) * (s - l3));
return area * 2 / l1;
*/
}
//向量的叉積
/* 計算向量的叉積(ABxAC A(x1,y1) B(x2,y2) C(x3,y3))是計算行列式
| x1-x0 y1-y0 |
| x2-x0 y2-y0 |
的結果(向量的叉積 AB X AC)
*/
//計算AB與AC的叉積---叉積的絕對值是兩向量所構成平行4邊形的面積
double crossProd(Point& pc) {
//計算ab X ac
return (pb.x - pa.x) * (pc.y - pa.y) - (pb.y - pa.y) * (pc.x - pa.x);
}
//判斷兩線段是不是相交
bool isIntersected(PartLine& pl) {
double d1, d2, d3, d4, d5, d6;
d1 = pl.crossProd(pb);
d2 = pl.crossProd(pa);
d3 = crossProd(pl.pa);
d4 = crossProd(pl.pb);
d5 = crossProd(pl.pa);
d6 = crossProd(pl.pb);
//printf("%f %f %f %f %f %f
", d1, d2, d3, d4, d5, d6);
bool cond1 = d1 * d2 <= 0, //pb和pa在pl的兩側或線段或線段的延長線上
cond2 = d3 * d4 <= 0, //pl.pa和pl.pb在this的兩側或線段或線段的延長線上
cond3 = d5 != 0, //pl.pa不在線段和延長線上
cond4 = d6 != 0; //pl.pb不在線段和延長線上
return cond1 && cond2 && cond3 && cond4;
}
//判斷兩線段是不是平行
bool isParallel(PartLine& pl) {
double v1 = crossProd(pl.pa),
v2 = crossProd(pl.pb);
return (v1 == v2) && (v1 != 0);
}
//沿pa點旋轉theta
PartLine rotateA(double theta) {
float nx = pa.x +(pb.x - pa.x) * cos(theta) - (pb.y - pa.y) * sin(theta),
ny = pa.y + (pb.x - pa.x) * sin(theta) + (pb.y - pa.y) * cos(theta);
return PartLine(pa, Point(nx, ny));
}
//沿pb點旋轉theta
PartLine rotateB(double theta) {
float nx = pb.x +(pa.x - pb.x) * cos(theta) - (pa.y - pb.y) * sin(theta),
ny = pb.y + (pa.x - pb.x) * sin(theta) + (pa.y - pb.y) * cos(theta);
return PartLine(Point(nx, ny), pb);
}
//判斷兩線段是不是堆疊或共線
bool inSameLine(PartLine& pl) {
double v1 = crossProd(pl.pa), v2 = crossProd(pl.pb);
if (v1 != v2) return false;
if (v1 != 0) return false;
return true;
}
//獲得兩線段的相交點---如果不相交返回valid=false
//如果多個交點,給出正告
Point getCrossPoint(PartLine& pl, bool& valid) {
valid = false;
if (!isIntersected(pl)) { //不相交
return Point();
}
if ( inSameLine(pl) ) { //有交點且共線
if ( pa.equal(pl.pa) ) { valid = true; return pa; }
if ( pa.equal(pl.pb) ) { valid = true; return pa; }
if ( pb.equal(pl.pa) ) { valid = true; return pb; }
if ( pb.equal(pl.pb) ) { valid = true; return pb; }
//多個焦點
cout << "毛病:計算交點結果數量為無窮" << endl;
valid = false;
return Point();
}
//相交
Point pt1, pt2, pt3, result;
pt1 = pa;
pt2 = pb;
pt3.x = (pt1.x + pt2.x) / 2;
pt3.y = (pt1.y + pt2.y) / 2;
double L1 = pl.crossProd(pt1), L2 = pl.crossProd(pt2), L3 = pl.crossProd(pt3);
printf("%f %f %f=%f
", L1, L2, L3, L1 + L2);
while(fabs(L1) > 1e⑺ || fabs(L2) > 1e⑺) {
valid = true;
if (fabs(L1) < fabs(L2))
pt2 = pt3;
else
pt1 = pt3;
pt3.x = (pt1.x + pt2.x) / 2;
pt3.y = (pt1.y + pt2.y) / 2;
result = pt3;
L1 = pl.crossProd(pt1), L2 = pl.crossProd(pt2), L3 = pl.crossProd(pt3);
printf("%f %f %f=%f
", L1, L2, L3, L1 - L2);
}
return pt3;
}
//獲得線段上離pt最近的點
Point getNearestPointToPoint(Point& pt) {
Point pt1, pt2, pt3, result;
pt1 = pa;
pt2 = pb;
pt3.x = (pt1.x + pt2.x) / 2;
pt3.y = (pt1.y + pt2.y) / 2;
double L1 = pt1.distance(pt), L2 = pt2.distance(pt), L3 = pt3.distance(pt);
if (L1 == L2) return pt3;
while(fabs(L1 - L2) > 1e⑺) {
if (L1 < L2)
pt2 = pt3;
else
pt1 = pt3;
pt3.x = (pt1.x + pt2.x) / 2;
pt3.y = (pt1.y + pt2.y) / 2;
result = pt3;
L1 = pt1.distance(pt);
L2 = pt2.distance(pt);
L3 = pt3.distance(pt);
//printf("%f %f %f=%f
", L1, L2, L3, L1 - L2);
}
return result;
}
//獲得1個點在線段上的鏡像點
Point getMirrorPoint(Point& pc) {
}
} PartLine;
int main(void)
{
Point p1(0, 0), p2(1, 1), p3(0, 1.1), p4(0.5, 0.5+1e⑴0), p5(0.5, 0.5⑴e⑴0), np;
PartLine pl1(p1, p2), pl2(p3, p4), pl3(p3, p5);
cout << pl1.getDistantToPoint(p3) << endl;
cout << "線段1和2相交?" << pl1.isIntersected(pl2) << endl;
np = pl1.getNearestPointToPoint(p5);
cout << "最近點:" << np.x << ", " << np.y << endl;
bool isvalid;
np = pl1.getCrossPoint(pl3, isvalid);
cout << "兩線段的相交點:" << (isvalid ? "有效":"無效") << "=" << np.x << ", " << np.y << endl;
PartLine plx = pl1.rotateA(M_PI / 2);
printf("旋轉90度后:%f %f %f %f
", plx.pa.x, plx.pa.y, plx.pb.x, plx.pb.y);
return 0;
}
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