poj 1408(计算几何)

1，求线段交点。

2，枚举各个面积

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <algorithm>

  5 #include <cmath>

  6 

  7 using namespace std;

  8 

  9 struct point

 10 {

 11     double x,y;

 12 };

 13 struct line

 14 {

 15     point a,b;

 16     line(){}

 17     line(point aa,point bb)

 18     {

 19         a=aa;

 20         b=bb;

 21     }

 22 };

 23 

 24 point intersection(line u,line v)

 25 {

 26     point ret=u.a;

 27     double t=((u.a.x-v.a.x) * (v.a.y-v.b.y) -(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));

 28     ret.x+=(u.b.x-u.a.x)*t;

 29     ret.y+=(u.b.y-u.a.y)*t;

 30     return ret;

 31 }

 32 

 33 double area_of_polygon(int n,point * p)

 34 {

 35     double s;

 36     if(n<3) return 0;

 37     s=p[0].y*(p[n-1].x-p[1].x);

 38     for(int i=1;i<n;i++)

 39         s+=p[i].y*(p[i-1].x-p[(i+1)%n].x);

 40     return fabs(s/2);

 41 }

 42 

 43 double a[35],b[35],c[35],d[35];

 44 int n;

 45 

 46 int main()

 47 {

 48     while(scanf("%d",&n))

 49     {

 50         if(n==0) break;

 51         a[0]=b[0]=c[0]=d[0]=0.0;

 52         a[n+1]=b[n+1]=c[n+1]=d[n+1]=1.0;

 53         for(int i=1;i<=n;i++)

 54             scanf("%lf",&a[i]);

 55         for(int i=1;i<=n;i++)

 56             scanf("%lf",&b[i]);

 57         for(int i=1;i<=n;i++)

 58             scanf("%lf",&c[i]);

 59         for(int i=1;i<=n;i++)

 60             scanf("%lf",&d[i]);

 61         sort(a,a+n+1);

 62         sort(b,b+n+1);

 63         sort(c,c+n+1);

 64         sort(d,d+n+1);

 65         double ans=0;

 66         point tmp[4];

 67         point intsec[35][35];

 68         for(int i=0;i<=n+1;i++)

 69         {

 70             point aa,bb;

 71             aa.x=a[i];

 72             aa.y=0;

 73             bb.x=b[i];

 74             bb.y=1;

 75             line u=line(aa,bb);

 76             //cout<<u.a.x<<" "<<u.a.y<<" "<<u.b.x<<" "<<u.b.y<<endl;

 77             for(int j=0;j<=n+1;j++)

 78             {

 79                 point cc,dd;

 80                 cc.x=0;

 81                 cc.y=c[j];

 82                 dd.x=1;

 83                 dd.y=d[j];

 84                 line v=line(cc,dd);

 85                 intsec[i][j]=intersection(u,v);

 86                 //cout<<intsec[i][j].x<<" "<<intsec[i][j].y<<endl;

 87             }

 88         }

 89         for(int i=1;i<=n+1;i++)

 90         {

 91             for(int j=1;j<=n+1;j++)

 92             {

 93                 tmp[0]=intsec[i-1][j];

 94                 tmp[1]=intsec[i-1][j-1];

 95                 tmp[2]=intsec[i][j-1];

 96                 tmp[3]=intsec[i][j];

 97                 double t=area_of_polygon(4,tmp);

 98                 if(t>ans)

 99                     ans=t;

100             }

101         }

102         printf("%.6f\n",ans);

103     }

104     return 0;

105 }