uva 11168

题意：给出平面上的n个点，求一条直线，使得所有点在该直线的同一侧且所有点到该直线的距离和最小，输出该距离和。

思路：要使所有点在该直线的同一侧，明显是直接利用凸包的边更优。所以枚举凸包的没条边，然后求距离和。直线一般式为Ax + By + C = 0.点(x0, y0)到直线的距离为 fabs(Ax0+By0+C)/sqrt(A*A+B*B).由于所有点在直线的同一侧，那么对于所有点，他们的(Ax0+By0+C)符号相同，显然可以累加出sumX和sumY，然后统一求和。

水。。。。

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 using namespace std;
 const double inf=0x3f3f3f3f;
 const int N = ;
 const double eps = *1e-;
 const double PI = acos(-1.0);

 double torad(double deg)
 {
     return deg/ * PI;
 }

 struct Point
 {
     double x, y;
     Point(double x=, double y=):x(x),y(y) { }
 };

 typedef Point Vector;

 Vector operator + (const Vector& A, const Vector& B)
 {
     return Vector(A.x+B.x, A.y+B.y);
 }
 Vector operator - (const Point& A, const Point& B)
 {
     return Vector(A.x-B.x, A.y-B.y);
 }
 double Cross(const Vector& A, const Vector& B)
 {
     return A.x*B.y - A.y*B.x;
 }

 Vector Rotate(const Vector& A, double rad)
 {
     return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
 }

 bool operator < (const Point& p1, const Point& p2)
 {
     return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
 }

 bool operator == (const Point& p1, const Point& p2)
 {
     return p1.x == p2.x && p1.y == p2.y;
 }

 // 点集凸包
 // 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
 // 如果不介意点集被修改，可以改成传递引用
 vector<Point> ConvexHull(vector<Point> p)
 {
     // 预处理，删除重复点
     sort(p.begin(), p.end());
     p.erase(unique(p.begin(), p.end()), p.end());

     int n = p.size();
     int m = ;
     vector<Point> ch(n+);
     for(int i = ; i < n; i++)
     {
         while(m >  && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-; i >= ; i--)
     {
         while(m > k && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     if(n > ) m--;
     ch.resize(m);
     return ch;
 }

 // 过两点p1, p2的直线一般方程ax+by+c=0
 // (x2-x1)(y-y1) = (y2-y1)(x-x1)
 void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double& b, double &c) {
   a = p2.y-p1.y;
   b = p1.x-p2.x;
   c = -a*p1.x - b*p1.y;
 }

 int main()
 {
     int t,n;
     double x,y;
     double sumx,sumy;
     double ans;
     int cas=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         vector<Point>p;
         sumx=;sumy=;
         for(int i=;i<n;i++)
         {
             scanf("%lf%lf",&x,&y);
             sumx+=x;sumy+=y;
             p.push_back(Point(x,y));
         }
         vector<Point>ch=ConvexHull(p);
         ans=inf;
         int edge=ch.size();
         if(edge<=)
         {
             ans=;
         }
         else
         {
             for(int i=;i<edge;i++)
             {
                 double a,b,c;
                 getLineGeneralEquation(ch[i],ch[(i+)%edge],a,b,c);
                 ans=min(ans,fabs(sumx*a+sumy*b+n*c)/sqrt(a*a+b*b));
             }
         }
         printf("Case #%d: %.3f\n",cas++,ans/n);
     }
     return ;
 }