题意：有n个点，问在一个m边形内哪个点与这n个点的距离平方和最小
题解：(ai-a0)^2=ai*ai+a0*a0-a*ai*a0
合起来就是a1*a1+...+an*an+n*a0*a0-2*a0*(a1+...+an)
取导数就是2*n*a0-2*a0*(a1+...+an)
可以知道在x y轴上各取n个点的平均值就是最小值
至于在不在m边形里，判断一下吧
不在里面就在线上，m条边求下导，求出最小值的位置
#include <iostream>
#include <fstream>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return ;
while (c != '-' && (c<'' || c>'')) c = getchar();
sgn = (c == '-') ? - : ;
ret = (c == '-') ? : (c - '');
while (c = getchar(), c >= ''&&c <= '') ret = ret * + (c - '');
ret *= sgn;
return ;
}
template <class T>
inline void pt(T x) {
if (x <) {
putchar('-');
x = -x;
}
if (x>) pt(x / );
putchar(x % + '');
}
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e5+;
const int inf = 1e9;
const double eps = 1e-;
struct Point{
double x, y;
Point(double a = , double b = ) :x(a), y(b){}
}a[N], b[N];
int n, m;
double cx, cy, C;
double good_x, good_y;
double cal(double x, double y){
double ans = ;
for (int i = ; i < n; i++)
ans += (a[i].x - x)*(a[i].x - x) + (a[i].y - y)*(a[i].y - y);
return ans;
}
double area(Point x, Point y, Point z){
return abs(x.x*y.y + y.x*z.y + z.x*x.y - x.x*z.y - y.x*x.y - z.x*y.y) / 2.0;
}
double work(Point x){
double ans = ;
for (int i = ; i < m; i++)
ans += area(x, b[i], b[(i + ) % m]);
return ans;
}
double papa(Point x){
return n*x.x*x.x + n*x.y*x.y - * x.x*cx - * x.y*cy;
}
Point cut(Point x, Point y, double k){
return Point(x.x + k*(y.x - x.x), x.y + k*(y.y - x.y));
}
double hehe(Point x, Point y){
double ans = min(papa(x), papa(y));
if (y.x != x.x){
double k = (y.y - x.y) / (y.x - x.x), b = x.y - k*x.x;
double _x = (k*cy + cx - n*k*b) / n / ( + k*k);
if (_x < min(x.x, y.x) || _x > max(x.x, y.x))return ans;
double _y = k*_x + b;
ans = min(ans, papa(Point(_x, _y)));
}
else {
if (min(x.y, y.y) <= good_y && good_y <= max(x.y, y.y))
ans = min(ans, papa(Point(x.x, good_y)));
}
return ans;
}
int main(){
rd(n);
cx = cy = C = ;
for (int i = ; i < n; i++){
rd(a[i].x), rd(a[i].y);
cx += a[i].x;
cy += a[i].y;
C += a[i].x*a[i].x + a[i].y*a[i].y;
}
rd(m);
for (int i = ; i < m; i++)rd(b[i].x), rd(b[i].y);
good_x = (double)cx / n;
good_y = (double)cy / n;
if (abs(work(b[]) - work(Point(good_x, good_y))) < eps)
printf("%.10f\n", cal(good_x, good_y));
else {
double ans = 1e19;
for (int i = ; i < m; i++)
ans = min(ans, hehe(b[i], b[( + i) % m]));
printf("%.10f\n", ans + C);
}
return ;
}