题目链接：点我点我
题意：给定n个点。
以下n行给出这n个点坐标。
给定m个点，以下m行给出这m个点坐标。
这m个点是一个凸包，顺时针给出的。
问：在凸包上随意找一个点(x, y) 使得这个点距离n个点的距离平方和最小。
问这个最小的距离平方和是多少。
思路：
首先化简一下公式。把变量(x,y)提出来会发现是一个简单的函数，且开口向上。所以有唯一解，解出这个(x,y) 记为 (good_x, good_y)
但这个点可能不是坐落在凸包内，若坐落在凸包外。则最优解一定是在凸包的边上，所以枚举每条边求个解就好了。
推断点在凸多边形内部用三角形面积相等就可以。
#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 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e5+10;
const int inf = 1e9;
const double eps = 1e-4;
struct Point{
double x, y;
Point(double a = 0, double b = 0) :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 = 0;
for (int i = 0; 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 = 0;
for (int i = 0; i < m; i++)
ans += area(x, b[i], b[(i + 1) % m]);
return ans;
}
double papa(Point x){
return n*x.x*x.x + n*x.y*x.y - 2 * x.x*cx - 2 * 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 / (1 + 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 = 0;
for (int i = 0; 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 = 0; 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[0]) - work(Point(good_x, good_y))) < eps)
printf("%.10f
", cal(good_x, good_y));
else {
double ans = 1e19;
for (int i = 0; i < m; i++)
ans = min(ans, hehe(b[i], b[(1 + i) % m]));
printf("%.10f
", ans + C);
}
return 0;
}