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  • uva 1396

    半平面的交,二分的方法;

      1 #include<cstdio>
      2 #include<algorithm>
      3 #include<cmath>
      4 #define eps 1e-6
      5 using namespace std;
      6 
      7 int dcmp(double x)
      8 {
      9     return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
     10 }
     11 
     12 struct Point
     13 {
     14     double x;
     15     double y;
     16     Point(double x = 0, double y = 0):x(x), y(y) {}
     17 };
     18 typedef Point Vector;
     19 
     20 Vector operator + (Point A, Point B)
     21 {
     22     return Vector(A.x + B.x, A.y + B.y);
     23 }
     24 
     25 Vector operator - (Point A, Point B)
     26 {
     27     return Vector(A.x - B.x, A.y - B.y);
     28 }
     29 
     30 Vector operator * (Point A, double p)
     31 {
     32     return Vector(A.x * p, A.y * p);
     33 }
     34 
     35 Vector operator / (Point A, double p)
     36 {
     37     return Vector(A.x / p, A.y / p);
     38 }
     39 double dot(Point a,Point b)
     40 {
     41     return a.x*b.x+a.y*b.y;
     42 }
     43 double cross(Point a,Point b)
     44 {
     45     return a.x*b.y-a.y*b.x;
     46 }
     47 
     48 Vector nomal(Vector a)
     49 {
     50     double l=sqrt(dot(a,a));
     51     return Vector(-a.y/l,a.x/l);
     52 }
     53 
     54 struct line
     55 {
     56     Point p;
     57     Vector v;
     58     double ang;
     59     line() {}
     60     line(Point p,Vector v):p(p),v(v)
     61     {
     62         ang=atan2(v.y,v.x);
     63     }
     64     bool operator<(const line &t)const
     65     {
     66         return ang<t.ang;
     67     }
     68 };
     69 
     70 bool onleft(line l,Point p)
     71 {
     72     return (cross(l.v,p-l.p)>0);
     73 }
     74 
     75 Point getintersection(line a,line b)
     76 {
     77     Vector u=a.p-b.p;
     78     double t=cross(b.v,u)/cross(a.v,b.v);
     79     return a.p+a.v*t;
     80 }
     81 
     82 int halfplanintersection(line *l,int n,Point *poly)
     83 {
     84     sort(l,l+n);
     85     int first,last;
     86     Point *p=new Point[n];
     87     line *q=new line[n];
     88     q[first=last=0]=l[0];
     89     for(int i=1; i<n; i++)
     90     {
     91         while(first<last && !onleft(l[i],p[last-1]))last--;
     92         while(first<last && !onleft(l[i],p[first]))first++;
     93         q[++last]=l[i];
     94         if(fabs(cross(q[last].v,q[last-1].v))<eps)
     95         {
     96             last--;
     97             if(onleft(q[last],l[i].p))q[last]=l[i];
     98         }
     99         if(first<last)p[last-1]=getintersection(q[last-1],q[last]);
    100     }
    101     while(first<last && !onleft(q[first],p[last-1]))last--;
    102     if((last-first )<=1)return 0;
    103     p[last]=getintersection(q[last],q[first]);
    104     int m=0;
    105     for(int i=first; i<=last; i++)poly[m++]=p[i];
    106     return m;
    107 }
    108 
    109 Point p[200],poly[200];
    110 line l[200];
    111 Point v[200],v2[200];
    112 int main()
    113 {
    114     int n,m;
    115     double x,y;
    116     while(scanf("%d",&n)&&n)
    117     {
    118         for(int i=0; i<n; i++)
    119         {
    120             scanf("%lf%lf",&x,&y);
    121             p[i]=Point(x,y);
    122         }
    123         for(int i=0; i<n; i++)
    124         {
    125             v[i]=p[(i+1)%n]-p[i];
    126             v2[i]=nomal(v[i]);
    127         }
    128         double left=0.0,right=20000.0;
    129         while(right-left>eps)
    130         {
    131             double mid=(left+right)/2;
    132             for(int i=0; i<n; i++)
    133                 l[i]=line(p[i]+v2[i]*mid,v[i]);
    134             m=halfplanintersection(l,n,poly);
    135             if(!m)right=mid;
    136             else left=mid;
    137         }
    138         printf("%.6lf
    ",left);
    139     }
    140     return 0;
    141 }
    View Code
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  • 原文地址:https://www.cnblogs.com/yours1103/p/3408373.html
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