今天打算做两道半平面交，一题卡太久了，心都碎了。。。
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#pragma warning(disable:4996)#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <cmath>#include <string>#include <algorithm>using
namespace std;#define maxn 2500#define eps 1e-7int
n;int
dcmp(double
x){ return
(x > eps) - (x < -eps);}struct
Point{ double
x, y; Point(){} Point(double
_x, double
_y) :x(_x), y(_y){} Point operator + (const
Point &b) const{ return
Point(x + b.x, y + b.y); } Point operator - (const
Point &b) const{ return
Point(x - b.x, y - b.y); } Point operator *(double
d) const{ return
Point(x*d, y*d); } Point operator /(double
d) const{ return
Point(x / d, y / d); } double
det(const
Point &b) const{ return
x*b.y - y*b.x; } double
dot(const
Point &b) const{ return
x*b.x + y*b.y; } Point rot90(){ return
Point(-y, x); } Point norm(){ double
len=sqrt(this->dot(*this)); return
Point(x, y) / len; } void
read(){ scanf("%lf%lf", &x, &y); }};#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))#define crossOp(p1,p2,p3) (dcmp(cross(p1,p2,p3)))Point isSS(Point p1, Point p2, Point q1, Point q2){ double
a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2); return
(p1*a2 + p2*a1) / (a1 + a2);}struct
Border{ Point p1, p2; double
alpha; void
setAlpha(){ alpha = atan2(p2.y - p1.y, p2.x - p1.x); }};bool
operator < (const
Border &a,const
Border &b) { int
c = dcmp(a.alpha - b.alpha); if
(c != 0) { return
c == 1; } else
{ return
crossOp(b.p1, b.p2, a.p1) > 0; }}bool
operator == (const
Border &a, const
Border &b){ return
dcmp(a.alpha - b.alpha) == 0;}Point isBorder(const
Border &a, const
Border &b){ return
isSS(a.p1, a.p2, b.p1, b.p2);}Border border[maxn];Border que[maxn];int
qh, qt;// check函数判断的是新加的半平面和由a,b两个半平面产生的交点的方向，若在半平面的左侧返回Truebool
check(const
Border &a, const
Border &b, const
Border &me){ Point is = isBorder(a, b); return
crossOp(me.p1, me.p2, is) > 0;}bool
convexIntersection(){ qh = qt = 0; sort(border, border + n); n = unique(border, border + n) - border; for
(int i = 0; i < n; i++){ Border cur = border[i]; while
(qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], cur)) --qt; while
(qh + 1 < qt&&!check(que[qh], que[qh + 1], cur)) ++qh; que[qt++] = cur; } while
(qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], que[qh])) --qt; while
(qh + 1 < qt&&!check(que[qh], que[qh + 1], que[qt - 1])) ++qh; return
qt - qh > 2;}Point ps[maxn];bool
judge(double
x){ for
(int i = 0; i < n; i++){ border[i].p1 = ps[i]; border[i].p2 = ps[(i + 1) % n]; } for
(int i = 0; i < n; i++){ Point vec = border[i].p2 - border[i].p1; vec=vec.rot90().norm(); vec = vec*x; border[i].p1 = border[i].p1 + vec; border[i].p2 = border[i].p2 + vec; border[i].setAlpha(); } return
convexIntersection();}int
main(){ while
(cin>>n&&n) { for
(int i = 0; i < n; i++){ ps[i].read(); } double
l=0, r=100000000; while
(dcmp(r-l)>0){ double
mid = (l + r) / 2; if
(judge(mid)) l = mid; else
r = mid; } printf("%.6lf\n", l); } return
0;} |
POJ3525 Most Distant Point from the Sea(半平面交),布布扣,bubuko.com
POJ3525 Most Distant Point from the Sea(半平面交)
原文：http://www.cnblogs.com/chanme/p/3710539.html