题意：
平面上有n个点，求一条直线使得所有点都在直线的同一侧。并求这些点到直线的距离之和的最小值。
分析：
只要直线不穿过凸包，就满足第一个条件。要使距离和最小，那直线一定在凸包的边上。所以求出凸包以后，枚举每个边求出所有点到直线的距离之和得到最小值。
点到直线距离公式为：
因为点都在直线同一侧，所以我们可以把加法“挪”到里面去，最后再求绝对值，所以可以预处理所有点的横坐标之和与纵坐标之和。当然常数C也要记得乘上n倍。
已知两点坐标求过该点直线的方程，这很好求不再赘述，考虑到直线没有斜率的情况，最终要把表达式中的分母乘过去。
1 //#define LOCAL
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7 using namespace std;
8
9 struct Point
10 {
11 double x, y;
12 Point(double x=0, double y=0):x(x), y(y) {}
13 };
14 typedef Point Vector;
15 Point operator + (Point A, Point B)
16 {
17 return Point(A.x+B.x, A.y+B.y);
18 }
19 Point operator - (Point A, Point B)
20 {
21 return Point(A.x-B.x, A.y-B.y);
22 }
23 bool operator < (const Point& A, const Point& B)
24 {
25 return A.x < B.x || (A.x == B.x && A.y < B.y);
26 }
27 bool operator == (const Point& A, const Point& B)
28 {
29 return A.x == B.x && A.y == B.y;
30 }
31 double Cross(Vector A, Vector B)
32 {
33 return A.x*B.y - A.y*B.x;
34 }
35
36 vector<Point> ConvexHull(vector<Point> p) {
37 // 预处理，删除重复点
38 sort(p.begin(), p.end());
39 p.erase(unique(p.begin(), p.end()), p.end());
40
41 int n = p.size();
42 int m = 0;
43 vector<Point> ch(n+1);
44 for(int i = 0; i < n; i++) {
45 while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
46 ch[m++] = p[i];
47 }
48 int k = m;
49 for(int i = n-2; i >= 0; i--) {
50 while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
51 ch[m++] = p[i];
52 }
53 if(n > 1) m--;
54 //for(int i = 0; i < m; ++i) printf("%lf %lf\n", ch[i].x, ch[i].y);
55 ch.resize(m);
56 return ch;
57 }
58
59 double sumx, sumy;
60
61 double Dist(Point a, Point b, int m)
62 {
63 double A = a.y-b.y, B = b.x-a.x, C = a.x*b.y - b.x*a.y;
64 //printf("%lf %lf", fabs(A*sumx+B*sumy+C), sqrt(A*A+B*B));
65 return (fabs(A*sumx+B*sumy+C*m) / sqrt(A*A+B*B));
66 }
67
68 int main(void)
69 {
70 #ifdef LOCAL
71 freopen("11168in.txt", "r", stdin);
72 #endif
73
74 int T;
75 scanf("%d", &T);
76 for(int kase = 1; kase <= T; ++kase)
77 {
78 int n;
79 vector<Point> p;
80 sumx = 0.0, sumy = 0.0;
81 scanf("%d", &n);
82 for(int i = 0; i < n; ++i)
83 {
84 double x, y;
85 scanf("%lf%lf", &x, &y);
86 p.push_back(Point(x, y));
87 sumx += x; sumy += y;
88 }
89 vector<Point> ch = ConvexHull(p);
90 int m = ch.size();
91 //for(int i = 0; i < m; ++i) printf("%lf %lf\n", ch[i].x, ch[i].y);
92 if(m <= 2)
93 {
94 printf("Case #%d: 0.000\n", kase);
95 continue;
96 }
97
98 double ans = 1e10;
99 for(int i = 0; i < m; ++i)
100 ans = min(ans, Dist(ch[i], ch[(i+1)%m], n));
101 printf("Case #%d: %.3lf\n", kase, ans/n);
102 }
103 }
来源：https://www.cnblogs.com/AOQNRMGYXLMV/p/4036163.html