UVA_11437_TriangleFun

11437 - Triangle Fun

Time limit: 1.000 seconds

In the picture below you can see a triangle ABC. Point D, E and F divides the sides BC, CA and

AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are

connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R.

So now a new triangle PQR is formed. Given triangle ABC your job is to nd the area of triangle

PQR.

Input

First line of the input le contains an integer N (0 < N < 1001) which denotes how many sets of

inputs are there. Input for each set contains six

oating-point number Ax, Ay, Bx, By, Cx, Cy.

(0  Ax; Ay; Bx; By; Cx; Cy  10000) in one line line. These six numbers denote that the coordinate of

points A, B and C are (Ax; Ay), (Bx; By) and (Cx; Cy) respectively. A, B and C will never be collinear.

Output

For each set of input produce one line of output. This one line contains an integer AREA. Here AREA

is the area of triangle PQR, rounded to the nearest integer.

Sample Input

2

3994.707 9251.677 4152.916 7157.810 5156.835 2551.972

6903.233 3540.932 5171.382 3708.015 213.959 2519.852

Sample Output

98099

206144

简单的计算几何题目

套点模板求个面积即可

#include <iostream>
#include <cstdio>
#include <cmath>               //需要用到部分函数
using namespace std;

const double PI=acos(-1.0);    //π的大小

const double eps=1e-8;         //允许误差
int dcmp(double x)              //误差修正
{
if(fabs(x)<eps)return 0;
if(x>0)        return 1;
return -1;
}
inline double sqr(double x)    //平方
{
return (x*x);
}

//以上基础

//以下点类

struct Point                                               //点类，向量类
{
double x,y;
Point(){}
Point(double a,double b):x(a),y(b){}
void input()
{
scanf("%lf%lf",&x,&y);
}
friend Point operator +(const Point &a,const Point &b)      //点之和
{
return Point(a.x+b.x,a.y+b.y);
}
friend Point operator -(const Point &a,const Point &b)      //点之差
{
return Point(a.x-b.x,a.y-b.y);
}
friend bool operator ==(const Point &a,const Point &b)      //点相同
{
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
friend Point operator *(const Point &a,const double &b)     //数乘
{
return Point(a.x*b,a.y*b);
}
friend Point operator *(const double &a,const Point &b)
{
return Point(a*b.x,a*b.y);
}
friend Point operator /(const Point &a,const double &b)      //除以某数
{
return Point(a.x/b,a.y/b);
}
double norm()                                  //向量的模长
{
return sqrt(sqr(x)+sqr(y));
}
};

typedef Point Vector;                              //向量是点的别名

double det(const Point &a,const Point &b)          //向量a，b的叉积
{                                                  //a转向b的有向面积（逆时针转为正）
return a.x*b.y-a.y*b.x;
}
double dot(const Point &a,const Point &b)          //向量a，b的点积
{
return a.x*b.x+a.y*b.y;
}

//以上点模板

//以下直线及线段模板

struct Segment                                     //线段类
{
Point a,b;
Segment(){}
Segment(Point x,Point y):a(x),b(y){}
Segment(double x1,double y1,double x2,double y2)
{
a.x=x1;a.y=y1;b.x=x2;b.y=y2;
}
};

typedef Segment Sline;                             //同时也是直线的两点表示法

bool parallel(const Sline &a,const Sline &b)  //线段所在直线是否平行
{
return !dcmp(det(a.a-a.b,b.a-b.b));
}

//求线段所在直线交点 如果平行返回0 否则返回1 交点保存在res
bool sline_make_point(const Sline &a,const Sline &b,Point &res)
{
if(parallel(a,b)) return 0;
double s1=det(a.a-b.a,b.b-b.a);
double s2=det(a.b-b.a,b.b-b.a);
res=(s1*a.b-s2*a.a)/(s1-s2);
return 1;
}

//以上线段类

//多边形类

const int MP=1000;
struct Polygon
{
int n;             //多边形的顶点数
Point a[MP];       //顶点
Polygon(){}
double area()                     //多边形面积
{
double sum=0;
a
=a[0];
for(int i=0;i<n;i++)
sum+=det(a[i+1],a[i]);    //利用叉积的有向面积性质
return sum/2.0;
}
};

int main()
{
int t;
Polygon AA,aa;
AA.n=3;aa.n=3;
Segment l1,l2,l3;
Point ab,bc,ca;
double ar;
scanf("%d",&t);
while(t--)
{
for(int i=0;i<3;i++)
AA.a[i].input();
ab=(AA.a[1]-AA.a[0])/3.0+AA.a[0];
bc=(AA.a[2]-AA.a[1])/3.0+AA.a[1];
ca=(AA.a[0]-AA.a[2])/3.0+AA.a[2];
l1=Segment(ab,AA.a[2]);
l2=Segment(bc,AA.a[0]);
l3=Segment(ca,AA.a[1]);
sline_make_point(l1,l2,aa.a[0]);
sline_make_point(l2,l3,aa.a[1]);
sline_make_point(l1,l3,aa.a[2]);
printf("%.0lf\n",fabs(aa.area()));
}
return 0;
}