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工作中遇到话流程图的项目,需要画带箭头的直线,经过摸索,解决;思路如下:
(1) 两个点(p1,p2)确定一个直线,以直线的一个端点(假设p2)为原点,设定一个角度
(2)以P2为原点得到向量P2P1(P),向量P旋转theta角得到向量P1,向量P旋转-theta角得到向量P2
(3)伸缩向量至制定长度,平移变量到直线的末端
(4)现在已经有3个点了,画线就可
具体代码如下:
void CworkflowDlg::DrawLine(CPoint p1, CPoint p2)
{
        CClientDC dc(this);//获取客户窗口DC
	CPen pen,pen1,*oldpen;
	int PenLineWidth=2;//为了根据线条宽度设置箭头的大小
	pen.CreatePen(PS_SOLID, PenLineWidth, RGB(0, 0, 0));
	pen1.CreatePen(PS_SOLID, PenLineWidth, RGB(0, 0, 0));
	oldpen=dc.SelectObject(&pen);
	
	double theta=3.1415926/15*PenLineWidth;//转换为弧度
	double Px,Py,P1x,P1y,P2x,P2y;
	//以P2为原点得到向量P2P1(P)
	Px=p1.x-p2.x;
	Py=p1.y-p2.y;
	//向量P旋转theta角得到向量P1
	P1x=Px*cos(theta)-Py*sin(theta);
	P1y=Px*sin(theta)+Py*cos(theta);
	//向量P旋转-theta角得到向量P2
	P2x=Px*cos(-theta)-Py*sin(-theta);
	P2y=Px*sin(-theta)+Py*cos(-theta);
	//伸缩向量至制定长度
	double x1,x2;
	int length=10;
	x1=sqrt(P1x*P1x+P1y*P1y);
	P1x=P1x*length/x1;
	P1y=P1y*length/x1;
	x2=sqrt(P2x*P2x+P2y*P2y);
	P2x=P2x*length/x2;
	P2y=P2y*length/x2;
	//平移变量到直线的末端
	P1x=P1x+p2.x;
	P1y=P1y+p2.y;
	P2x=P2x+p2.x;
	P2y=P2y+p2.y;
	
	dc.MoveTo(p1.x,p1.y);
	dc.LineTo(p2.x,p2.y);
	dc.SelectObject(&pen1);
	dc.MoveTo(p2.x,p2.y);
	dc.LineTo(P1x,P1y);
	dc.MoveTo(p2.x,p2.y);
	dc.LineTo(P2x,P2y);
	dc.MoveTo(P1x,P1y);
	dc.LineTo(P2x,P2y);
	CPoint ptVertex[3];
	ptVertex[0].x = p2.x;
	ptVertex[0].y = p2.y;
	ptVertex[1].x = P1x;
	ptVertex[1].y = P1y;
	ptVertex[2].x = P2x;
	ptVertex[2].y = P2y;
        //填充三角形区域
	CBrush br(RGB(40,130,170));  
	CRgn rgn; 
	rgn.CreatePolygonRgn(ptVertex,3,ALTERNATE);
	dc.FillRgn(&rgn, &br);  
	dc.SelectObject(oldpen);
	br.DeleteObject();  
	rgn.DeleteObject();
}  具体效果如下:
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原文地址:http://blog.csdn.net/monaso/article/details/51353696