HDU 3694 Fermat Point in Quadrangle)

Here are four points for you to find the minimum distance between one point and four points.

The feathed point of the Quadrilateral: the convex side is the intersection of two diagonal lines, and the concave side is a concave point.

PS:

The horse points of the triangle:

1. If the three inner angles of a triangle are less than 120 °, the three inner angles of the triangle are exactly the same as those of the third equi-fenma point, that is, the angle of the triangle on which the vertex is located is equal to 120 °. Therefore, the fenma point of a triangle is also called the Equi-angle center of a triangle.

2. If a triangle has an inner angle greater than or equal to 120 °, the vertices of the inner angle are the points of distance and minimum.

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          Using namespace std; const int kind = 26; const int maxn = 250*1000; // Note RE, word length * Number of words const int M = 5100000; struct Point {double x, y ;}; // less than 0, indicating that the polar angle of the vector p0p1 is greater than the Polar Angle of p0p2, double multiply (Point p1, Point p2, Point p0) {return (p1.x-forwarded X) * (p2.y-p0.y)-(p2.x-p0.x) * (p1.y-p0.y);} double dis (Point p1, Point p2) {return (sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y);} void Graham_scan (Point PointSet [], Point ch [], int n, int & len) {int I, j, k = 0, top = 2; Point tmp; // locate the lowest and leftmost Point for (I = 1; I
         
           0) | (multiply (PointSet [j], PointSet [k], PointSet [0]) = 0) & (dis (PointSet [0], pointSet [j])
          
            = 0) top --; // The current vertex and all vertices in the stack meet the left-facing relationship. ch [++ top] = PointSet [I];} len = top + 1;} Point intersection (Point u1, Point u2, Point v1, Point v2) {Point ret = u1; double t = (u1.x-v1.x) * (v1.y-v2.y)-(u1.y-v1.y) * (v1.x-v2.x )) /(u1.x-u2.x) * (v1.y-v2.y)-(u1.y-u2.y) * (v1.x-v2.x); ret. x + = (u2.x-u1.x) * t; ret. y + = (u2.y-u1.y) * t; return ret;} Point p [4], ch [4], point; int len; int main () {int x1, y1, x2, y2, x3, y3, x4, y4; int x [4], y [ 4]; while (scanf ("% lf", & p [0]. x, & p [0]. y) {int flag = 0; if (p [0]. x! =-1 | p [0]. y! =-1) flag = 1; for (int I = 1; I <4; I ++) {scanf ("% lf", & p [I]. x, & p [I]. y); if (p [I]. x! =-1 | p [I]. y! =-1) flag = 1;} if (! Flag) break; double minn = 10000000, ans = 0.0; int k = 0; Graham_scan (p, ch, 4, len); for (int I = 0; I <4; I ++) {ans = 0.0; for (int j = 0; j <4; j ++) {ans + = dis (p [I], p [j]);} if (ans