前幾天的練習(xí)賽有一道計(jì)算幾何題，一向討厭計(jì)算幾何的我推了一下之后就沒做了。
后來比賽結(jié)束的時(shí)候發(fā)現(xiàn)他們都過了，后悔不已。故做了一下，求三角形外接圓圓心那個(gè)我使用
垂直平分線相交的那個(gè)做的。上次他們說有公式，我在書上找了個(gè)圓心公式，可是代進(jìn)去不對。
估計(jì)是書上公式寫錯(cuò)了...
Bounding box
Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 28 Accepted Runs: 14 Multiple test files
The Archeologists of the Current Millenium (ACM) now and then discover
ancient artifacts located at vertices of regular polygons. The
moving sand dunes of the desert render the excavations difficult and
thus once three vertices of a polygon are discovered there is a need to cover
the entire polygon with protective fabric.
Input contains multiple cases. Each case describes one polygon.
It starts with an integer
n ≤ 50, the number of vertices in the polygon, followed
by three pairs of real numbers giving the x and y
coordinates of three vertices of the polygon. The numbers are separated
by whitespace. The input ends with a n equal 0, this case should
not be processed.
For each line of input, output one line in the format shown below, giving
the smallest area of a rectangle which can cover all the vertices of the
polygon and whose sides are parallel to the x and y axes.
Sample input
4
10.00000 0.00000
0.00000 -10.00000
-10.00000 0.00000
6
22.23086 0.42320
-4.87328 11.92822
1.76914 27.57680
23
156.71567 -13.63236
139.03195 -22.04236
137.96925 -11.70517
0
Output for the sample input
Polygon 1: 400.000
Polygon 2: 1056.172
Polygon 3: 397.673
// solution by alpc12
#include <cstdio>
#include <cmath>
const double EPS = 1e-8;
const double PI = acos(-1.0);
const double INF = 1e100;
#define Min(a, b) (a<b?a:b)
#define Max(a, b) (a>b?a:b)
struct Point {
double x;
double y;
Point() {}
Point(double xx, double yy) {
x = xx;
y = yy;
}
};
struct Line {
double a, b, c;
Point st, end;
Line() {}
Line(Point& u, Point& v) {
st = u;
end = v;
a = v.y - u.y;
b = u.x - v.x;
c = a*u.x + b*u.y;
}
};
#define sqr(a) ((a)*(a))
#define dist(a, b) (sqrt( sqr((a).x-(b).x)+sqr((a).y-(b).y) ))
#define cross(a, b, c) (((b).x-(a).x)*((c).y-(a).y)-((b).y-(a).y)*((c).x-(a).x))
inline int dblcmp(double a, double b = 0.0) {
if(fabs(a-b) < EPS) return 0;
return a < b ? -1 : 1;
}
Line bisector(Point& a, Point& b) {
Line line(a, b), ans;
double midx = (a.x+b.x)/2, midy = (a.y+b.y)/2;
ans.a = -line.b, ans.b = line.a, ans.c = ans.a*midx + ans.b*midy;
return ans;
}
int line_line_intersect(Line& l1, Line& l2, Point& s) {
double det = l1.a * l2.b - l2.a * l1.b;
if(dblcmp(det) == 0) {
return -1;
}
s.x = (l2.b*l1.c - l1.b*l2.c) / det;
s.y = (l1.a*l2.c - l2.a*l1.c) / det;
return 1;
}
int center_3point(Point& a, Point& b, Point& c, Point& s, double& r) {
Line x = bisector(a, b), y = bisector(b, c);
if(line_line_intersect(x, y, s) == 1) {
r = dist(s, a);
return 1;
}
return 0;
}
Point p[55];
int main() {
//freopen("t.in", "r", stdin);
int i, n, tc = 0;
Point cent;
while(scanf("%d", &n), n) {
for(i = 0; i < 3; ++i) scanf("%lf %lf ", &p[i].x, &p[i].y);
double r;
if(center_3point(p[0], p[1], p[2], cent, r) == 1) {
for(i = 0; i < 3; ++i)
p[i].x -= cent.x, p[i].y -= cent.y;
}
double alpha = acos(p[0].x / r);
double theta = 2 * PI / n;
double xmin = INF, xmax = -INF, ymin = INF, ymax = -INF;
for(i = 0; i < n; ++i) {
p[i] = Point(r * cos(alpha + i * theta),
r * sin(alpha + i * theta));
xmin = Min(xmin, p[i].x);
xmax = Max(xmax, p[i].x);
ymin = Min(ymin, p[i].y);
ymax = Max(ymax, p[i].y);
}
printf("Polygon %d: %.3lf\n", ++tc, (xmax-xmin)*(ymax-ymin));
}
return 0;
}