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C++判断矩形相交的方法

来源:本站原创|时间:2020-01-10|栏目:C语言|点击:

本文实例讲述了C++判断矩形相交的方法。分享给大家供大家参考。具体如下:

已知2矩形原点和宽高,判断2矩形相交,相交矩形

相交判断原理:

假定矩形是用一对点表达的(minx, miny) (maxx, maxy),那么两个矩形
    rect1{(minx1, miny1)(maxx1, maxy1)}
    rect2{(minx2, miny2)(maxx2, maxy2)} 
相交的结果一定是个矩形,构成这个相交矩形rect{(minx, miny) (maxx, maxy)}的点对坐标是: 
    minx   =   max(minx1,   minx2) 
    miny   =   max(miny1,   miny2) 
    maxx   =   min(maxx1,   maxx2) 
    maxy   =   min(maxy1,   maxy2) 
如果两个矩形不相交,那么计算得到的点对坐标必然满足: 
  ( minx  >  maxx ) 或者 ( miny  >  maxy )
判定是否相交,以及相交矩形是什么都可以用这个方法一体计算完成
 
设计3个类:

1. 点类:x,y
2. 矩形类:点,宽,高
3. 判断相交类

程序实现:

CPoint.h 
#import <Foundation/Foundation.h>
@interface CPoint : NSObject 
{
  int x; //点坐标
  int y;
}
-(void) print;
-(void) setX: (int) vx;
-(void) setY: (int) vy;
-(void) setXY:(int) vx :(int) vy;
-(int) x;
-(int) y;
@end 
CPoint.m
#import "CPoint.h"
@implementation CPoint
-(void) print
{
  NSLog(@"the point is (%i, %i)",x,y);
}
-(void) setX: (int) vx
{
  x = vx;
}
-(void) setY: (int) vy
{
  y = vy;
}
-(void) setXY:(int)vx :(int)vy
{
  x = vx;
  y = vy;
}
-(int) x
{
  return x;
}
-(int) y
{
  return y;
}
@end
CRect.h
#import <Foundation/Foundation.h>
#import "CPoint.h"
@interface CRect : NSObject
{
  int w; //矩形长
  int h; //矩形高
}
-(void) print;
-(int) w;
-(int) h;
-(void) setW: (int) vw;
-(void) setH: (int) vh;
-(void) setWH: (int) vw :(int) vh;
-(CPoint *) origin;
-(void) setOrigin: (CPoint *) pt;
@end
CRect.m
#import "CRect.h"
@implementation CRect
{
  CPoint *origin; //点
}
-(void) print
{
  NSLog(@"the rect:(x:%i, y:%i, w:%i,h:%i)",origin.x, origin.y, w, h);
}
-(int) w
{
  return w;
}
-(int) h
{
  return h;
}
-(void) setW:(int)vw
{
  w = vw;
}
-(void) setH:(int)vh
{
  h = vh;
}
-(void) setWH:(int)vw :(int)vh
{
  w = vw;
  h = vh;
}
-(CPoint *) origin
{
  return origin;
}
-(void) setOrigin:(CPoint *)pt
{
  origin = pt;
}
@end
DoCRect.h
#import <Foundation/Foundation.h>
#import "CRect.h"
@interface DoCRect : NSObject
-(BOOL) isIntersect:(CRect *) rect1 :(CRect *) rect2; //矩形相交否
-(CRect *) intersectRect: (CRect *) rect1 :(CRect *) rect2; //相交矩形
@end
DoCRect.m
#import "DoCRect.h"
@implementation DoCRect
//矩形是否相交
-(BOOL) isIntersect:(CRect *)rect1 :(CRect *)rect2
{
  int minx = MAX(rect1.origin.x, rect2.origin.x);
  int miny = MAX(rect1.origin.y, rect2.origin.y);
  int maxx = MIN(rect1.origin.x+rect1.w, rect2.origin.x+rect2.w);
  int maxy = MIN(rect1.origin.y+rect1.h, rect2.origin.y+rect2.h);
  if (minx>maxx || miny>maxy)
  {
    return NO;
  }
  return YES;
}
-(CRect *) intersectRect:(CRect *)rect1 :(CRect *)rect2
{
  int minx = MAX(rect1.origin.x, rect2.origin.x);
  int miny = MAX(rect1.origin.y, rect2.origin.y);
  int maxx = MIN(rect1.origin.x+rect1.w, rect2.origin.x+rect2.w);
  int maxy = MIN(rect1.origin.y+rect1.h, rect2.origin.y+rect2.h); 
  CRect * rect = [[CRect alloc] init];
  CPoint *p = [[CPoint alloc] init];
  if (NO == [self isIntersect:rect1 :rect2])//no isIntersect
  {
    [p setXY:minx :miny];
    [rect setOrigin:p];
    rect.w = 0;
    rect.h = 0;
    return rect;
  }
  [p setXY:minx :miny];
  [rect setOrigin:p];
  rect.w = ABS(maxx-minx);
  rect.h = ABS(maxy - miny);
  return rect; 
}
@end
main.m 测试
#import <Foundation/Foundation.h>
#import "DoCRect.h"
int main(int argc, const char * argv[])
{
  @autoreleasepool
  {
    NSLog(@"Hello,判断矩形相交,返回矩形的原点和长高");
    //初始化对象
    CRect *myrect1 = [[CRect alloc] init];
    CRect *myrect2 = [[CRect alloc] init];
    CPoint *p1 = [[CPoint alloc] init];
    CPoint *p2 = [[CPoint alloc] init];
    DoCRect *dorect = [[DoCRect alloc] init];
    //原点变量
    [p1 setXY:200 :420];
    [p2 setXY:400 :300];
    //设置矩形原点
    [myrect1 setOrigin:p1];
    [myrect1 setWH:250 :75];
    [myrect1 print];
    [myrect2 setOrigin:p2];
    [myrect2 setWH:100 :180];
    [myrect2 print];
    //判断2矩形是否相交
    BOOL insersect = [dorect isIntersect:myrect1 :myrect1];
    NSLog(@" two rect is :%@",insersect?@"YES":@"NO");
    //返回相交矩形
    //CRect *inserectRect = [[CRect alloc] init];
    CRect *inserectRect = [dorect intersectRect:myrect1 :myrect2];
    [inserectRect print];
  }
  return 0;
}

希望本文所述对大家的C++程序设计有所帮助。

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