UVALive 5102 Fermat Point in Quadrangle polar sorting + finding the Point closest to the four points of two-dimensional coordinates, uvalivefermat

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Author: User

UVALive 5102 Fermat Point in Quadrangle polar sorting + finding the Point closest to the four points of two-dimensional coordinates, uvalivefermat

Question:

Four points on two-dimensional coordinates

Q:

Find a vertex to minimize the distance between the vertex and the four vertices.

Output distance and.

Ideas:

If the four vertices are not convex and have four edges. It must be the best endpoint.

Otherwise, the intersection of two diagonal lines is the best, which can be proved simply.

For a 4-sided convex shape, the first polar angle is sorted.

`#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef double ll;const int N = 5;int n = 4;double x[N], y[N];struct Point {    ll x, y, dis;} s[4], p0;ll Dis(ll x1, ll y1, ll x2, ll y2){    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);}int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb){    double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y);    if (delta<0.0) return 1;    else if (delta==0.0) return 0;    else return -1;}bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1){    int type=Cmp_PolarAngel(p3, p1, p2);    if (type<0) return true;    return false;}int Cmp(const void*p1, const void*p2){    struct Point*a1=(struct Point*)p1;    struct Point*a2=(struct Point*)p2;    int type=Cmp_PolarAngel(*a1, *a2, p0);    if (type<0) return -1;    else if (type==0) {        if (a1->dis<a2->dis) return -1;        else if (a1->dis==a2->dis) return 0;        else return 1;    }    else return 1;}double cal(double x1, double y1, double x2, double y2) {return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}int main() {while (~scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x[0], &y[0], &x[1], &y[1], &x[2], &y[2], &x[3], &y[3])) {if(x[0] == -1) break;double ans = 1e10, di;for(int i = 0; i < n; i ++) {di = 0.0;for(int j = 0; j < n; j ++) {if(j == i) continue;di += cal(x[i], y[i], x[j], y[j]);}ans = min(ans, di);}double xx = x[0] + x[1] + x[2] + x[3];double yy = y[0] + y[1] + y[2] + y[3];di = 0.0;for(int j = 0; j < n; j ++) {di += cal(xx/4, yy/4, x[j], y[j]);}ans = min(ans, di);p0.x = x[0], p0.y = y[0];for(int i = 0; i < 4; i ++) {s[i].x = x[i];s[i].y = y[i];}for(int i = 0; i < n; i ++) {s[i].dis = cal(s[0].x, s[0].y, s[i].x, s[i].y);}qsort(s+1, n-1, sizeof(struct Point), Cmp);x[0] = s[0].x; y[0] = s[0].y;x[1] = s[2].x; y[1] = s[2].y;x[2] = s[1].x; y[2] = s[1].y;x[3] = s[3].x; y[3] = s[3].y;double k1 = (y[0] - y[1]) / (x[0] - x[1]);double k2 = (y[3] - y[2]) / (x[3] - x[2]);double ansx, ansy;if(x[0] == x[1]) {ansx = x[0];if(x[2] == x[3]) {ansy = yy / 4;} else {ansy = k2 * (ansx - x[2]) + y[2];}} else {if(x[2] == x[3]) {ansx = x[2];ansy = k1 * (ansx - x[1]) + y[1];} else {if(k1 != k2) {ansx = (y[2] - y[1] +  k1*x[1] - k2*x[2]) / (k1 - k2);ansy = k1*(ansx - x[1]) + y[1];} else {ansx = 1000;ansy = 1000;}}}di = 0.0;for(int j = 0; j < n; j ++) {di += cal(ansx, ansy, x[j], y[j]);}ans = min(ans, di);printf("%.4f\n", ans);}return 0;}`

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