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;
}
简单几何(数学公式+凸包) UVA 11168 Airport
原文:http://www.cnblogs.com/Running-Time/p/4953133.html