2017ACM/ICPC亚洲区沈阳站 C Hdu-6219 Empty Convex Polygons 计算几何 最大空凸包

题面

题意:给你一堆点,求一个最大面积的空凸包,里面没有点.

题解:红书板子,照抄完事,因为题目给的都是整点,所以最后答案一定是.5或者.0结尾,不用对答案多做处理

 #include<bits/stdc++.h>
 #define N 55
 using namespace std;
 struct rec
 {
     double x,y;
 };
 rec operator -(rec a,rec b)
 {
     rec c;
     c.x=a.x-b.x;
     c.y=a.y-b.y;
     return c;
 }
 double sqr(double a)
 {
     return a*a;
 }
 int sign(double a)
 {
     if (fabs(a) <= 1e-) return ;
     return a<?- :;
 }
 bool operator <(rec a,rec b)
 {
      return sign(b.y-a.y)> || sign(b.y-a.y)== && sign(b.x-a.x)>;
 }
 double max(double a,double b)
 {
     return a>b ?a:b;
 }
 double length(rec a)
 {
     return sqrt(sqr(a.x)+sqr(a.y));
 }
 double cross(rec a,rec b)
 {
     return a.x*b.y-a.y*b.x;
 }
 rec dot[N],lis[N];
 double opt[N][N];
 int seq[N],n,len;
 double ans;
 bool Compare(rec a,rec b)
 {
     int temp=sign(cross(a,b));
     if (temp!=) return temp>;
     temp=sign(length(b)-length(a));
     return temp>;
 }
 void solve(int vv)
 {
     int t,i,j,_len;
     for (i=len=;i<n;i++)
         if (dot[vv]<dot[i]) lis[len++]=dot[i]-dot[vv];
     for (int i=;i<len;i++)
         for (int j=;j<len;j++)
             opt[i][j]=;
     sort(lis,lis+len,Compare);
     double v;
     for (t=;t<len;t++)
     {
         _len=;
         for (i=t-;i>= && sign(cross(lis[t],lis[i])) ==   ;i--);
         while (i>=)
         {
             v=cross(lis[i],lis[t])/;
             seq[_len++]=i;
             for (j=i-; j>= && sign(cross(lis[i]-lis[t], lis[j]-lis[t])) > ;j--);
             if (j>=) v+=opt[i][j];
             ans=max(ans,v);
             opt[t][i]=v;
             i=j;
         }
         for (i = _len-;i>=;i--)
             opt[t][seq[i]]=max(opt[t][seq[i]],opt[t][seq[i+]]);
     }
 }
 int T;
 int main()
 {
     scanf("%d",&T);
     while (T--)
     {
         scanf("%d",&n);
         for (int i=;i<n;i++) scanf("%lf%lf",&dot[i].x,&dot[i].y);
         ans=;
         for (int i=;i<n;i++) solve(i);
         printf("%.1lf\n",ans);
     }
     return ;
 }