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CGL_3_A: Area

Area

For a given polygon g, computes the area of the polygon.

g is represented by a sequence of points $p_1$, $p_2$,…, $p_n$ where line segments connecting pi and pi+1 (1 ≤ in−1) are sides of g. The line segment connecting pn and p1 is also a side of the polygon.

Note that the polygon is not necessarily convex.

Input

The input consists of coordinates of the points p1,…, pn in the following format:

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2
3
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5
n
x1 y1
x2 y2
:
xn yn

The first integer n is the number of points. The coordinate of a point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them.

Output

Print the area of the polygon in a line. The area should be printed with one digit to the right of the decimal point.

Constraints

  • 3 ≤ n ≤ 100
  • -10000 ≤ $x_i,y_i$ ≤ 10000
  • No point will occur more than once.
  • Two sides can intersect only at a common endpoint.

Sample Input 1

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2
3
4
3
0 0
2 2
-1 1

Sample Output 1

1
2.0

Sample Input 2

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2
3
4
5
4
0 0
1 1
1 2
0 2

Sample Output 2

1
1.5

Source: https://onlinejudge.u-aizu.ac.jp/problems/CGL_3_A

求多边形的面积

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#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
struct Point{
double x,y;
Point(){}
Point(double x,double y):x(x),y(y){}
};
typedef Point Vector;
Vector operator + (Vector A,Vector B){
return Vector(A.x+B.x,A.y+B.y);
}
Vector operator - (Vector A,Vector B){
return Vector(A.x-B.x,A.y-B.y);
}
Vector operator * (Vector A,double p){
return Vector(A.x*p,A.y*p);
}
Vector operator / (Vector A,double p){
return Vector(A.x/p,A.y/p);
}
bool operator < (const Point& a,const Point& b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
const double eps = 1e-10;
int dcmp(double x){
if(fabs(x)<eps) return 0; else return (x<0?-1:1);
}
bool operator == (const Point& a,const Point& b){
return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
//点积
double Dot(Vector A,Vector B){
return A.x*B.x+A.y*B.y;
}
double Length(Vector A){
return sqrt(Dot(A,A));
}
double Angle(Vector A,Vector B){
return acos(Dot(A,B)/Length(A)/Length(B));
}

//叉积
double Cross(Vector A,Vector B){
return (A.x*B.y-A.y*B.x);
}
//三角形面积的二倍的叉乘公式
double Area2(Point A,Point B,Point C){
return Cross(B-A,C-A);
}
//向量旋转,rad是弧度
Vector Rotate(Vector A,double rad){
return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
//计算向量的单位法线,先逆时针旋转90度,然后把长度归一化
Vector Normal(Vector A){
double Len=Length(A);
return Vector(-A.y/Len,A.x/Len);
}
//求两直线交点
//调用前请确保P+tv和Q+tw有唯一交点,当且仅当Cross(v,w)非0
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){
Vector u=P-Q;
double t=Cross(w,u)/Cross(v,w);
return P+v*t;
}
//点到直线的距离
double DistanceToLine(Point P,Point A,Point B){
Vector v1=B-A,v2=P-A;
return fabs(Cross(v1,v2))/Length(v1);
}
//点到线段的距离
double DistanceToSegment(Point P,Point A,Point B){
if(A==B) return Length(P-A);
Vector v1=B-A,v2=P-A,v3=P-B;
if(dcmp(Dot(v1,v2))<0) return Length(v2);
else if(dcmp(Dot(v1,v3))>0) return Length(v3);
else return fabs(Cross(v1,v2))/Length(v1);
}
//点在直线上的投影
Point GetLineProjection(Point P,Point A,Point B){
Vector v=B-A;
return A+v*(Dot(v,P-A)/Dot(v,v));
}

const int maxn=100+10;
double PolygonArea(Point* p,int n){
double area=0;
for(int i=1;i<n-1;i++){
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
}
return area/2;
}
int main(){
int n;
cin>>n;
Point Pg[maxn];
for(int i=0;i<n;i++){
cin>>Pg[i].x>>Pg[i].y;
}
cout<<fixed<<setprecision(1)<<PolygonArea(Pg,n)<<"\n";

}