解題區/解題區 - UVa

UVa 10173 - Smallest Bounding Rectangle

contents

  1. 1. Problem
  2. 2. Input
  3. 3. Output
  4. 4. Sample Input
  5. 5. Sample Output
  6. 6. Solution

Problem

Given the Cartesian coordinates of n (> 0) 2-dimensional points, write a program that computes the area of their smallest bounding rectangle (smallest rectangle containing all the given points).

Input

The input file may contain multiple test cases. Each test case begins with a line containing a positive integer n (< 1001) indicating the number of points in this test case. Then follows n lines each containing two real numbers giving respectively the x- and y-coordinates of a point. The input terminates with a test case containing a value 0 for n which must not be processed.

Output

For each test case in the input print a line containing the area of the smallest bounding rectangle rounded to the 4th digit after the decimal point.

Sample Input

1
2
3
4
5
6
7
8
9
10
3
-3.000 5.000
7.000 9.000
17.000 5.000
4
10.000 10.000
10.000 20.000
20.000 20.000
20.000 10.000
0

Sample Output

1
2
80.0000
100.0000

Solution

題目描述：

給予平面上 N 個點，找到一個最小矩形覆蓋所有點座標。

題目解法：

做一次單調鏈，得到逆時針的凸包順序，接著第一步找到四邊平行 XY 軸的最小矩形。

接著對於卡邊四邊上的頂點，找到矩形邊與凸包邊夾角最小的頂點，進行向量的旋轉，然後再進行計算矩形的長寬。

對於矩形的長寬，直接套用點線投影公式。

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#include <stdio.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define eps 1e-8
struct Pt {
    double x, y;
    Pt(double a = 0, double b = 0):
    	x(a), y(b) {}
    bool operator<(const Pt &a) const {
        if(fabs(x-a.x) > eps)	return x < a.x;
        return y < a.y;
    }
    bool operator==(const Pt &a) const {
        return fabs(x-a.x) < eps && fabs(y-a.y) < eps;
    }
    Pt operator+(const Pt &a) const {
        return Pt(x + a.x, y + a.y);
    }
    Pt operator-(const Pt &a) const {
        return Pt(x - a.x, y - a.y);
    }
    Pt operator/(const double val) const {
        return Pt(x / val, y / val);
    }
    Pt operator*(const double val) const {
        return Pt(x * val, y * val);
    }
};
typedef Pt Vector;
double dist(Pt a, Pt b) {
    return hypot(a.x - b.x, a.y - b.y);
}
double dot(Pt a, Pt b) {
    return a.x * b.x + a.y * b.y;
}
double cross2(Pt a, Pt b) {
    return a.x * b.y - a.y * b.x;
}
double cross(Pt o, Pt a, Pt b) {
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
int between(Pt a, Pt b, Pt c) {
    return dot(c - a, b - a) >= 0 && dot(c - b, a - b) >= 0;
}
int onSeg(Pt a, Pt b, Pt c) {
    return between(a, b, c) && fabs(cross(a, b, c)) < eps;
}
double distProjection(Pt as, Pt at, Pt s) {
    double a, b, c;
    a = at.y - as.y;
    b = as.x - at.x;
    c = - (a * as.x + b * as.y);
    return fabs(a * s.x + b * s.y + c) / hypot(a, b);
}
struct Seg {
    Pt s, e;
};
int calcIntersection(Seg a, Seg b, Pt &p) {
    double a1, b1, c1, a2, b2, c2;
    double d, dx, dy;
    a1 = a.s.y-a.e.y, b1 = -a.s.x+a.e.x;
    a2 = b.s.y-b.e.y, b2 = -b.s.x+b.e.x;
    c1 = a1*a.s.x + b1*a.s.y;
    c2 = a2*b.s.x + b2*b.s.y;
    d = a1*b2 - a2*b1;
    dx = c1*b2 - c2*b1;
    dy = a1*c2 - a2*c1;
    if(fabs(d) < eps) // NONE or LINE
        return 0;
    p.x = dx / d, p.y = dy / d;
    /*printf("%lf %lf - %lf %lf\n", a.s.x, a.s.y, a.e.x, a.e.y);
    printf("%lf %lf - %lf %lf\n", b.s.x, b.s.y, b.e.x, b.e.y);
    printf("%lf %lf\n", p.x, p.y);*/
    return onSeg(a.s, a.e, p) && onSeg(b.s, b.e, p);
}
int inPolygon(Pt p[], int n, Pt q) {
    int i, j, cnt = 0;
    for(i = 0, j = n-1; i < n; j = i++) {
        if(p[i].y > q.y != p[j].y > q.y &&
           q.x < (p[j].x-p[i].x)*(q.y-p[i].y)/(p[j].y-p[i].y) + p[i].x)
           cnt++;
    }
    return cnt&1;
}
double calcArea(Pt p[], int n) {
    if(n < 3)   return 0.0;
    double ret = 0;
    int i;
    p[n] = p[0];
    for(i = 0; i < n; i++)
        ret += p[i].x * p[i+1].y - p[i].y * p[i+1].x;
    return fabs(ret)/2;
}
int monotone(int n, Pt p[], Pt ch[]) {
    sort(p, p+n);
    int i, m = 0, t;
    for(i = 0; i < n; i++) {
        while(m >= 2 && cross(ch[m-2], ch[m-1], p[i]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    for(i = n-1, t = m+1; i >= 0; i--) {
        while(m >= t && cross(ch[m-2], ch[m-1], p[i]) <= 0)
            m--;
        ch[m++] = p[i];
    }
    return m-1;
}
#define INF 1e+30 
double smallRect(int n, Pt ch[]) {
    double lx, ly, rx, ry;
    int up, down, left, right;
    double ret = INF;
    lx = ly = INF;
    rx = ry = -INF;
    
    up = down = left = right = 0;
    for(int i = 0; i < n; i++) {
        if(ch[i].x > rx)	rx = ch[i].x, right = i;
        if(ch[i].y > ry)	ry = ch[i].y, up = i;
        if(ch[i].x < lx)	lx = ch[i].x, left = i;
        if(ch[i].y < ly)	ly = ch[i].y, down = i;
    }
    
    int corner[] = {up, down, left, right};
    Pt vec[] = {Pt(-1, 0), Pt(1, 0), Pt(0, -1), Pt(0, 1)};
    
    ret = (rx - lx) * (ry - ly);
    for(int j = 0; j < 4; j++) {
        while(true) {
            Pt a = ch[corner[j]], b = ch[(corner[j]+1)%n], c = vec[j];
            if(fabs(cross2(b - a, c)) < eps)
                corner[j] = (corner[j] + 1)%n;
            else
                break;
        }
    }
//	for(int j = 0; j < 4; j++) {
//		Pt a = ch[corner[j]], b = vec[j];
//		printf("Pt[%lf %lf], Vector[%lf %lf]\n", a.x, a.y, b.x, b.y);
//	}	
    for(int i = 0; i < n; i++) {
        double mxVal = -INF, cos, sin;
        int mxIdx = 0;
        for(int j = 0; j < 4; j++) {
            Pt a = ch[corner[j]], b = ch[(corner[j]+1)%n], c = vec[j];
            double cosA = dot(b - a, c) / dist(b, a) / dist(c, Pt(0, 0));
            if(mxVal < cosA)
                mxVal = cosA, mxIdx = j;
//			printf("cos %lf\n", cosA);
        }
        cos = mxVal, sin = sqrt(1 - cos * cos);
//		printf("sin %lf cos %lf\n", sin, cos);
        for(int j = 0; j < 4; j++) {
            double tx, ty;
            tx = vec[j].x * cos - vec[j].y * sin;
            ty = vec[j].x * sin + vec[j].y * cos;
            vec[j] = Pt(tx, ty);
//			printf("%lf %lf\n", tx, ty);
        }
//		for(int j = 0; j < 4; j++) {
//			Pt a = ch[corner[j]], b = vec[j];
//			printf("Pt[%lf %lf], Vector[%lf %lf]\n", a.x, a.y, b.x, b.y);
//		}		
        for(int j = 0; j < 4; j++) {
            while(true) {
                Pt a = ch[corner[j]], b = ch[(corner[j]+1)%n], c = vec[j];
                if(fabs(cross2(b - a, c)) < eps)
                    corner[j] = (corner[j] + 1)%n;
                else
                    break;
            }
        }
        double w = distProjection(ch[corner[0]], ch[corner[0]]+vec[0], ch[corner[1]]);
        double h = distProjection(ch[corner[2]], ch[corner[2]]+vec[2], ch[corner[3]]);
//		printf("w %lf h %lf area %lf\n\n", w, h, w * h);
        ret = min(ret, w * h);
    }
    return ret;
}
int main() {
    Pt p[2048], ch[2048];
    int n, m;
    int testcase, cases = 0;
    while(scanf("%d", &n) == 1 && n) {
    	for(int i = 0; i < n; i++)
    		scanf("%lf %lf", &p[i].x, &p[i].y);
        m = monotone(n, p, ch);
//        for(int i = 0; i < m; i++)
//        	printf("%lf %lf\n", ch[i].x, ch[i].y);
    	
    	if(m < 3) {
    		printf("%.4lf\n", 0);
    		continue;
    	}
        double ret = smallRect(m, ch);
        printf("%.4lf\n", ret);
    }
    return 0;
}