简单几何(数学公式+凸包) UVA 11168 Airport

题目传送门

题意：找一条直线，使得其余的点都在直线的同一侧，而且使得到直线的平均距离最短。

分析：训练指南P274，先求凸包，如果每条边都算一边的话，是O (n ^ 2)，然而根据公式知直线一般式为Ax + By + C = 0.点(x0, y0)到直线的距离为：fabs(Ax0+By0+C)/sqrt(A*A+B*B)。

所以只要先求出x的和以及y的和，能在O (1)计算所有距离和。

两点式直线方程p1 (x1, y1)，p2 (x2, y2）转换成一般式直线方程：A = y1 - y2, B = x2 - x1, C = -A * x1 - B * y1;

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/10 星期二 11:24:09
* File Name     :UVA_11168.cpp
 ************************************************/

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point    {
    double x, y;
    Point () {}
    Point (double x, double y) : x (x), y (y) {}
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    Point operator * (double p) const {       //向量乘以标量
        return Point (x * p, y * p);
    }
    Point operator / (double p) const {       //向量除以标量
        return Point (x / p, y / p);
    }
    Point operator + (const Point &r) const {
        return Point (x + r.x, y + r.y);
    }
    bool operator < (const Point &r) const {
        return x < r.x || (x == r.x && y < r.y);
    }
    bool operator == (const Point &r) const {
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
};
typedef Point Vector;
Point read_point(void)  {
    double x, y;    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
double length(Vector V) {
    return sqrt (dot (V, V));
}
vector<Point> convex_hull(vector<Point> ps) {
    sort (ps.begin (), ps.end ());
    ps.erase (unique (ps.begin (), ps.end ()), ps.end ());
    int n = ps.size (), k = 0;
    vector<Point> qs (n * 2);
    for (int i=0; i<n; ++i) {
        while (k > 1 && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-2]) <= 0)    k--;
        qs[k++] = ps[i];
    }
    for (int t=k, i=n-2; i>=0; --i) {
        while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-2]) <= 0)    k--;
        qs[k++] = ps[i];
    }
    qs.resize (k-1);
    return qs;
}

double sqr(double x)    {
    return x * x;
}

int main(void)    {
    int T, cas = 0;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        vector<Point> ps;
        double x, y, sx = 0, sy = 0;
        for (int i=0; i<n; ++i) {
            scanf ("%lf%lf", &x, &y);
            sx += x;    sy += y;
            ps.push_back (Point (x, y));
        }
        vector<Point> qs = convex_hull (ps);
        if ((int) qs.size () <= 2)  {
            printf ("Case #%d: %.3f\n", ++cas, 0.0);    continue;
        }
        double mn = 1e9;
        qs.push_back (qs[0]);
        for (int i=0; i<qs.size ()-1; ++i)    {
            double A = qs[i].y - qs[i+1].y;
            double B = qs[i+1].x - qs[i].x;
            double C = -A * qs[i].x - B * qs[i].y;
            double tmp = fabs (A * sx + B * sy + n * C) / sqrt (sqr (A) + sqr (B));
            if (mn > tmp)   mn = tmp;
        }
        printf ("Case #%d: %.3f\n", ++cas, mn / n);
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}