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#include "stdafx.h"
#include "BranchPoint.h"
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
using namespace std;
// 简单的路径信息,路径规划的直接结果
struct Path
{
vector<string> wirelist; // 路径中的线缆名
vector<P> points; // 路径上的点,包括卡箍和分支点
vector<int> inOut; // 卡箍方向是正向为0,反向为1,分支点为-1
double dia; // 路径直径
Path()
{
dia = 0;
}
// 返回path中点的数量
inline int size()
{
assert(inOut.size() == points.size());
return points.size();
}
// 把路径中经过的卡箍和分支点都标记成使用过的状态
void 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 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 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 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 postProcess()
{
this->fixBranchPoint();
// this->processDir();
this->markPoint();
}
// 翻转整条路径
void 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();
}
}
}
};