两个凸多边形之间的切线

给定两个凸多边形，我们需要找到这些多边形的上下切线。
如下图所示， $T_{RL}$和$T_{LR}$分别显示上切线和下切线。

例子：

Input : First Polygon  : {(2, 2), (3, 3), (5, 2), (4, 0), (3, 1)} 
        Second Polygon : {(-1, 0), (0, 1), (1, 0), (0, -2)}.
Output : Upper Tangent - line joining (0,1) and (3,3)
         Lower Tangent - line joining (0,-2) and (4,0)

概述：
如图所示，我们有两个凸多边形，

为了找到上切线，我们从两点开始。 a的最右点和b的最左点。连接它们的线标记为1。这条线穿过多边形b(不在多边形b上方)，因此我们沿b的逆时针下一个点进行标记，该线标记为2。现在，该线在多边形b上方， 美好的！但是该线与多边形a相交，因此我们移至顺时针的下一个点，在图片中标记为3。它再次与多边形a相交，因此我们移至线4。这条线与b相交，因此我们移至线5。因此，这是给定多边形的上切线。
为了找到较低的切线，我们需要反向移动多边形，即，如果线与多边形b相交，则移动到下一个顺时针方向，如果线与多边形a相交，则移动到下一个逆时针方向。

上切线算法：

L Algorithm for lower tangent:
L // C++ program to find upper tangent of two polygons.#includeusing namespace std;  // stores the center of polygon (It is made// global becuase it is used in compare function)pair mid;  // determines the quadrant of a point// (used in compare())int quad(pair p){    if (p.first >= 0 && p.second >= 0)        return 1;    if (p.first <= 0 && p.second >= 0)        return 2;    if (p.first <= 0 && p.second <= 0)        return 3;    return 4;}  // Checks whether the line is crossing the polygonint orientation(pair a, pair b,                                 pair c){    int res = (b.second-a.second)*(c.first-b.first) -              (c.second-b.second)*(b.first-a.first);      if (res == 0)        return 0;    if (res > 0)        return 1;    return -1;}  // compare function for sortingbool compare(pair p1, pair q1){    pair p = make_pair(p1.first - mid.first,                                p1.second - mid.second);    pair q = make_pair(q1.first - mid.first,                                q1.second - mid.second);      int one = quad(p);    int two = quad(q);      if (one != two)        return (one < two);    return (p.second*q.first < q.second*p.first);}  // Finds upper tangent of two polygons 'a' and 'b'// represented as two vectors.void findUpperTangent(vector > a,                      vector > b){    // n1 -> number of points in polygon a    // n2 -> number of points in polygon b    int n1 = a.size(), n2 = b.size();      // To find a point inside the convex polygon(centroid),    // we sum up all the coordinates and then divide  by    // n(number of points). But this would be a floating-point    // value. So to get rid of this we multiply points    // initially with n1 and then find the centre and    // then divided it by n1 again.    // Similarly we do divide and multiply for n2 (i.e.,    // elements of b)      // maxa and minb are used to check if polygon a    // is left of b.    int maxa = INT_MIN;    for (int i=0; i rightmost point of a    int ia = 0, ib = 0;    for (int i=1; i a[ia].first)            ia = i;      // ib -> leftmost point of b    for (int i=1; i 0)            inda = (inda + 1) % n1;          while (orientation(a[inda], b[indb], b[(n2+indb-1)%n2]) < 0)        {            indb = (n2+indb-1)%n2;            done = 0;        }    }      cout << "upper tangent (" << a[inda].first << ","        << a[inda].second << ") (" << b[indb].first        << "," << b[indb].second << ")\n";}  // Driver codeint main(){    vector > a;    a.push_back({2, 2});    a.push_back({3, 1});    a.push_back({3, 3});    a.push_back({5, 2});    a.push_back({4, 0});      vector > b;    b.push_back({0, 1});    b.push_back({1, 0});    b.push_back({0, -2});    b.push_back({-1, 0});      findUpperTangent(a, b);      return 0;}Output:
Upper tangent (0,1) (3,3)
Note that the above code only finds upper tangent. We can similarly find lower tangent.