#include "pch.h" // use stdafx.h in Visual Studio 2017 and earlier
#include "Path.h"

ClipSet clipSet;
BranchPointSet branchPointSet;

P dirP(P &A)
{
	P B;
	B.x = A.dx;
	B.y = A.dy;
	B.z = A.dz;
	return B;
}

Path::Path()
{
	dia = 0;
}

// 返回path中点的数量
int Path::size()
{
	assert(inOut.size() == points.size());
	return points.size();
}

void Path::markPoint()
{
	for (int i = 0; i < points.size(); i++)
	{
		P &tp = points[i];
		if (tp.type == 1 || tp.type == 2)
		{
			BranchPoint *pb = branchPointSet.getBranchPointPointer(tp.ref);
			pb->used = true;
			pb->pc1->used = true;
			pb->pc2->used = true;
		}
		else if (tp.type == 0)
		{
			clipSet.c[tp.ref].used = true;
		}
	}
}

// 固定所有经过的分支点
void Path::fixBranchPoint()
{
	for (int i = 0; i < points.size(); i++)
	{
		P &tp = points[i];
		if (tp.type == 1)
		{
			BranchPoint *pb = branchPointSet.getBranchPointPointer(tp.ref);
			pb->fix();
		}
	}
}

// 检查路径合法性
void Path::illegalCheck()
{
	map<P, int> vis;
	for (int i = 0; i < points.size(); i++)
	{
		assert(vis[points[i]] == 0);
		vis[points[i]] = 1;
		if (i == 0 || i == points.size() - 1)
		{
			assert(points[i].type == 3);
		}
		if (points[i].type != 0)
		{
			assert(inOut[i] < 0);
		}
	}
}

// 将路径中的点翻转成正常反向，并生成inOut数组
void Path::processDir()
{
	for (int j = 0; j < points.size(); j++)
	{
		if (points[j].type != 0)
		{
			inOut.push_back(-1);
			continue;
		}
		double reverse = 0;
		P dir = dirP(points[j]);
		if (j != 0)
		{
			reverse += pi / 2 - Angle(points[j] - points[j - 1], dir);
		}
		if (j != points.size() - 1)
		{
			reverse += pi / 2 - Angle(points[j + 1] - points[j], dir);
		}
		if (reverse < 0)
		{
			// points[j].reverse();
			inOut.push_back(1);
		}
		else
			inOut.push_back(0);
	}
}

// 在路径确定后，处理一些能在pass层面处理的信息
void Path::postProcess()
{
	this->fixBranchPoint();
	// this->processDir();
	this->markPoint();
}

// 翻转整条路径
void Path::reverse()
{
	std::reverse(points.begin(), points.end());
	assert(inOut.size() == points.size());
	for (int i = 0; i < points.size(); i++)
	{
		if (inOut[i] != -1)
		{
			inOut[i] = inOut[i] ^ 1;
			points[i].reverse();
		}
	}
}