ICPCCamp2017 Day 4 F Factory（三分套三分）

大体题意：

给你二维坐标面上n个点，让你求出一个点，到这n个点的距离和最小？

思路：

赛后才想出怎么做来= =

写一写表达式：

sqrt((x0-x1)^2 + (y0-y1)^2 ) + sqrt((x0-x2)^2 + (y0-y2)^2 ) + sqrt((x0-x3)^2 + (y0-y3)^2 ) ....

观察发现，x0是一个凹凸函数（二次函数）关系，y0也是一个凹凸函数（二次函数）关系，  加起来还是一个二次函数关系（凹凸函数）。

那么两个未知量，都是凹凸性，直接三分x中在三分y即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define sqr(x) ((x)*(x))
using namespace std;
const double eps = 1e-10;
int n, T;
int dcmp(double a,double b){
    if (fabs(a-b) < eps) return 0;
    if (a < b) return -1;
    return 1;
}
struct Node{
    double x,y;
    void read(){
        scanf("%lf %lf",&x, &y);
    }
}p[1007];
double judge(double x,double y){
    double sum = 0;
    for (int i = 0; i < n; ++i){
        double xx = p[i].x;
        double yy = p[i].y;
        sum += sqrt(sqr(x-xx) + sqr(y-yy));
    }
    return sum;
}
const int oo = 1e6+7;
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for (int i = 0; i < n; ++i) p[i].read();
        double lx = -oo, rx = oo;
        double ly = -oo, ry = oo;
        for (int i = 0; i < 100; ++i){
            ly = -oo, ry = oo;
            double mx1 = (rx+lx) / 2;
            double mx2 = (mx1 + rx) / 2;
            for (int j = 0; j < 100; ++j){
                double my1 = (ly + ry) / 2;
                double my2 = (my1 + ry) / 2;
                double t1 = judge(mx1,my1);
                double t2 = judge(mx1,my2);
                if (dcmp(t1,t2) == -1) ry = my2;
                else ly = my1;
            }
            double ny1 = (ly+ry) / 2;
            ly = -oo, ry = oo;
            for (int j = 0; j < 100; ++j){
                double my1 = (ly + ry) / 2;
                double my2 = (my1 + ry) / 2;
                double t1 = judge(mx2,my1);
                double t2 = judge(mx2,my2);
                if (dcmp(t1,t2) == -1) ry = my2;
                else ly = my1;
            }
            double ny2 = (ly+ry) / 2;
            double nt1 = judge(mx1,ny1), nt2 = judge(mx2,ny2);
            if (dcmp(nt1,nt2) == -1) rx = mx2;
            else lx = mx1;
        }
        printf("%.10f %.10f\n",(lx+rx)/2,(ly+ry)/2);
    }
    return 0;
}
/**
1
3
-3 0
0 3
3 0

**/