codeforces 394E Lightbulb for Minister 简单几何

题目链接：点我点我
题意：给定n个点。下面n行给出这n个点坐标。
给定m个点，下面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：在凸包上任意找一个点(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。
判断点在凸多边形内部用三角形面积相等即可。

#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\n", 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\n", ans + C);    }    return 0;}