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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
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原文地址:http://www.cnblogs.com/Running-Time/p/4953133.html