poj3348 Cows 凸包+多边形面积 水题
/*
poj3348 Cows 凸包+多边形面积 水题
floor向下取整，返回的是double
*/
#include<stdio.h>
#include<math.h>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
struct point
{
double x,y;
};
int n;
point dian[10000+10],zhan[10000+10];
//////////////////////////////////////////////////
point *mo_dian;
double mo_distance(point p1,point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double mo_xmult(point p2,point p0,point p1)//p1在p2左返回负，在右边返回正
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool mo_ee(double x,double y)
{
double ret=x-y;
if(ret<0) ret=-ret;
if(ret<eps) return 1;
return 0;
}
bool mo_gg(double x,double y) { return x > y + eps;} // x > y
bool mo_ll(double x,double y) { return x < y - eps;} // x < y
bool mo_ge(double x,double y) { return x > y - eps;} // x >= y
bool mo_le(double x,double y) { return x < y + eps;} // x <= y
bool mo_cmp(point a,point b) // 第一次排序
{
if( mo_ee(a.y ,b.y ) )
return mo_ll(a.x, b.x);
return mo_ll(a.y,b.y);
}
bool mo_cmp1(point a,point b) // 第二次排序
{
double len = mo_xmult(b,mo_dian[0],a);
if( mo_ee(len,0.0) )
return mo_ll(mo_distance(mo_dian[0],a),mo_distance(mo_dian[0],b));
return mo_gg(len,0.0);
}
int mo_graham(int n,point *dian,point *stk)
{
int i,top=1;
mo_dian=dian;
sort(mo_dian,mo_dian+n,mo_cmp);
sort(mo_dian+1,mo_dian+n,mo_cmp1);
stk[0]=mo_dian[0];
stk[1]=mo_dian[1];
for(i=2;i<n;++i)
{
while(top>0&&mo_xmult(mo_dian[i],stk[top-1],stk[top])<=0) --top;
stk[++top]=mo_dian[i];
}
return top+1;
}
double mo_area_polygon(point *dian,int n)
{
int i;
point yuan;
yuan.x=yuan.y=0;
double ret=0;
for(i=0;i<n;++i)
{
ret+=mo_xmult(dian[(i+1)%n],yuan,dian[i]);
}
if(ret<0) ret=-ret;
return ret/2;
}
////////
int main()
{
int i;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;++i)
{
scanf("%lf%lf",&dian[i].x,&dian[i].y);
}
int ret=mo_graham(n,dian,zhan);
double area=mo_area_polygon(zhan,ret);
printf("%.0lf\n",floor(area/50));
}
return 0;
}