hdu2036

Posted 2021-01-21 yinwuxiao

题解：

求多边形面积

分成很多块三角形求就可以了

凹的也是支持的

代码：

 

#include <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
const double eps=1e-8;
struct Point
{
  double x,y;
  Point(double x1,double y1) {x=x1,y=y1;}
  Point(){};
  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);
  }
  double operator *(const Point b) const
  {
    return x*b.x+y*b.y;
  }
  double operator ^(const Point b) const
  {
    return x*b.y-y*b.x;
  }
  Point operator *(double k)
  {
    return Point(x*k,y*k);
  }
  Point operator /(double k)
  {
    return Point(x/k,y/k);
  }
  bool operator ==(Point b)
  {
    return b.x==x&&b.y==y?1:0;
  } 
};
struct Line
{
  Point x,y;
  Line() {};
  Line(Point x1,Point y1){x=x1,y=y1;};
};
double lenth(Point x)
{
  return sqrt(x.x*x.x+x.y*x.y);
}
double angle(Point x,Point y)
{
  return acos(x*y/lenth(x)/lenth(y));
}
int dcmp(double x)
{
  if (x<-eps) return(-1);
  else if (x>eps) return(1);
  else return(0);
}
Point rotate(Point x,double r)
{
  return Point(x.x*cos(r)-x.y*sin(r),x.x*sin(r)+x.y*cos(r));
}
Point gtp(Line x,Line y)
{
  Point v1=x.y-x.x; Point v2=y.y-y.x;
  double k=((x.x^v1)-(y.x^v1))/(v2^v1);
  return y.x+v2*k;
}
double distance(Point x,Line y)
{
  Point p1=x-y.x,p2=y.y-y.x;
  return fabs((p1^p2)/lenth(p2));
}
const int N=1000;
Point p[N];
int main()
{
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  ios::sync_with_stdio(false);
  int n;
  while (cin>>n&&n)
  {
    rep(i,1,n)
    {
      int x,y;
      cin>>x>>y;
      p[i]=Point(x,y);
    }
    double ans=0;
    rep(i,2,n-1)
      ans+=(p[i]-p[1])^(p[i+1]-p[1]);
    ans/=2;
    printf("%.1lf
",ans);
  }
  return 0;
}