Learning Algorithm: Computational Geometry and seek to find the convex hull points dotted line personally

 

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Pre-knowledge:
Computational Geometry foundation

 

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Find the convex hull :

vector<P> convex(vector<P> l)
{
    vector<P> ans, s;
    P tmp(lim, lim);
    int pos = 0;
    for (int i = 0; i < l.size(); i++)
        if (l[i] < tmp)
            tmp = l[i], pos = i;
    for (int i = pos, cnt_ = 0; cnt_ < l.size(); cnt_++, i = (i + 1) % l.size())
    {
        while (s.size() >= 2 && sgn(cross(s[s.size() - 2], s[s.size() - 1], l[i])) <= 0)
            s.pop_back();
        s.push_back(l[i]);
        pos = i;
    }
    ans = s; s.clear();
    for (int i = pos, cnt_ = 0; cnt_ < l.size(); cnt_++, i = (i - 1) % l.size())
    {
        while (s.size() >= 2 && sgn(cross(s[s.size() - 2], s[s.size() - 1], l[i])) <= 0)
            s.pop_back();
        s.push_back(l[i]);
        pos = i;
    }
    for (int i = 1; i + 1 < s.size(); i++)
        ans.push_back(s[i]);
    return ans;
}

Find the convex hull

 

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Seeking whether the points within the polygon

By calculating the number of sides is equal on both sides of the point, if they are equal then the points within the polygon

Meanwhile, the point is calculated on a straight line, find the point where the straight line

bool P_In_S(P a, vector<P> b)
{
    int cnt = b.size();
    int cnt_ = 0;
    b.push_back(b[0]);
    for (int i = 0; i < cnt; i++)
    {
        if (onL(a, L(b[i], b[i + 1])))    return true;
        int k = sgn(cross(a, b[i], b[i + 1]));
        int d1 = sgn(a.y - b[i].y),
            d2 = sgn(a.y - b[i + 1].y);
        if (k < 0 && d1 >= 0 && d2 < 0)cnt_++;
        if (k > 0 && d2 >= 0 && d1 < 0)cnt_--;
    }
    return cnt_ != 0;
}

Point is within the polygon

 

Seeking two vectors intersect

bool L_is_Inter(L a, L b)
{
    int c1 = sgn(cross(a.s, a.t, b.s)),
        c2 = sgn(cross(a.s, a.t, b.t)),
        c3 = sgn(cross(b.s, b.t, a.s)),
        c4 = sgn(cross(b.s, b.t, a.t));
    if (c1*c2 < 0 && c3*c4 < 0)return true;
    if (!c1  &&  onL(b.s, a))return true;
    if (!c2  &&  onL(b.t, a))return true;
    if (!c3  &&  onL(a.s, a))return true;
    if (!c4  &&  onL(a.t, a))return true;
    return false;
}

Two line segments intersect

 

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Through the above two knowledge points:

Possible to find two polygons intersect

bool IF_inter(vector<P> a, vector<P> b)
{
    //printf("1\n");
    for (int i = 0; i < a.size(); i++)
        if (P_In_S(a[i], b))
            return true;
    //printf("2\n");
    for (int i = 0; i < b.size(); i++)
        if (P_In_S(b[i], a))
            return true;
    int cnt1 = a.size(), cnt2 = b.size();
    a.push_back(a[0]), b.push_back(b[0]);
    //printf("3\n");
    for (int i = 0; i < cnt1; i++)
        for (int j = 0; j < cnt2; j++)
            if (L_is_Inter(L(a[i], a[i + 1]), L(b[j], b[j + 1])))
                return true;
    return false;
}

Two polygons intersect