题目大意：按顺序给出一个多边形的顶点，求这个多边形内核的面积。答案保留两位输出。

思路：半平面交。加边的时候要讨论一下第一个点和最后一个点，否则会wa的很惨。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1510
#define EPS 1e-8
#define DCMP(a) (fabs(a) < EPS)
using namespace std;

struct Point{
	double x,y;

	Point(double _ = .0,double __ = .0):x(_),y(__) {}	
	Point operator +(const Point &a)const {
		return Point(x + a.x,y + a.y);
	}
	Point operator -(const Point &a)const {
		return Point(x - a.x,y - a.y);
	}
	Point operator *(double a)const {
		return Point(x * a,y * a);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}point[MAX],p[MAX],ploygen[MAX];
struct Line{
	Point p,v;
	double alpha;

	Line(Point _,Point __):p(_),v(__) {
		alpha = atan2(v.y,v.x);
	}
	Line() {}	
	bool operator <(const Line &a)const {
		return alpha < a.alpha;
	}
}line[MAX],q[MAX];

int cases;
int points,lines;

inline void Initialize();
inline void MakeLine(const Point &a,const Point &b);
inline double Cross(const Point &a,const Point &b);
inline bool OnLeft(const Line &l,const Point &p);

inline Point GetIntersection(const Line &a,const Line &b);
inline double HalfPlaneIntersection();
inline double GetArea(int cnt);

int main()
{
	for(cin >> cases;cases; --cases) {
		scanf("%d",&points);
		Initialize();
		for(int i = 1;i <= points; ++i)
			point[i].Read();
		for(int i = points;i > 1; --i)
			MakeLine(point[i],point[i - 1]);
		MakeLine(point[1],point[points]);
		sort(line + 1,line + lines + 1);
		printf("%.2lf\n",HalfPlaneIntersection());
	}
	return 0;
}

inline void Initialize()
{
	lines = 0;
}

inline void MakeLine(const Point &a,const Point &b)
{
	line[++lines] = Line(a,b - a);
}

inline double Cross(const Point &a,const Point &b)
{
	return a.x * b.y - a.y * b.x;
}

inline Point GetIntersection(const Line &a,const Line &b)
{
	Point u = a.p - b.p;
	double temp = Cross(b.v,u) / Cross(a.v,b.v);
	return a.p + a.v * temp;
}

inline bool OnLeft(const Line &l,const Point &p)
{
	return Cross(l.v,p - l.p) > 0;
}

inline double HalfPlaneIntersection()
{
	int front = 1,tail = 1;
	q[tail] = line[1];
	for(int i = 2;i <= lines; ++i) {
		while(front < tail && !OnLeft(line[i],p[tail - 1]))	--tail;
		while(front < tail && !OnLeft(line[i],p[front]))	++front;
		if(DCMP(Cross(line[i].v,q[tail].v)))
			q[tail] = OnLeft(q[tail],line[i].p) ? line[i]:q[tail];
		else	q[++tail] = line[i];
		if(front < tail)	p[tail - 1] = GetIntersection(q[tail],q[tail - 1]);
	}
	while(front < tail && !OnLeft(q[front],p[tail - 1]))	--tail;
	if(front == tail)	return .0;
	p[tail] = GetIntersection(q[front],q[tail]);
	int cnt = 0;
	for(int i = front;i <= tail; ++i)
		ploygen[++cnt] = p[i];
	return GetArea(cnt);
}

inline double GetArea(int cnt)
{
	double re = Cross(ploygen[cnt],ploygen[1]);
	for(int i = 1;i < cnt; ++i)
		re += Cross(ploygen[i],ploygen[i + 1]);
	return fabs(re / 2);
}

POJ 1279 Art Gallery 多边形内核面积