POJ2451 Uyuw's Concert(半平面交)

题意就是给你很多个半平面，求半平面交出来的凸包的面积。

半平面交有O(n^2)的算法，就是每次用一个新的半平面去切已有的凸包，更新，这个写起来感觉也不是特别好写。

另外一个O(nlogn)的算法是将半平面交极角排序，然后用一个双端队列去维护半平面交，每次加入一个半平面，根据之前的交点的位置退掉半平面，方法跟凸包非常相像，（不同的是加入队列的时候还要考虑加入半平面会使队首的平面变得无效，因为会有两个while.最后还要考虑最后加入的半平面绕了一圈之后使得队首的半平面变得无效。）其实这里有点不太懂后面再慢慢补一补吧。

下面的代码用的是CLJ的模板，看大神写的模板能学习到很多东西。我对于计算几何还是不太熟悉，因为如何利用cross求很多东西其实都不太熟，例如如何用cross求两直线的交点。还有里面的border的极角排序的<重载，unique去重时需要的==重载。更高深的原理日后慢慢研究。。－ －0

#pragma warning(disable:4996)

#include <iostream>

#include <cstring>

#include <cstdio>

#include <vector>

#include <cmath>

#include <string>

#include <algorithm>

using namespace std;



#define maxn 25000

#define eps 1e-8



int n;



int dcmp(double x){

	return (x > eps) - (x < -eps);

}



struct Point

{

	double x, y;

	Point(){}

	Point(double _x, double _y) :x(_x), y(_y){}

	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);

	}

	Point operator *(double d) const{

		return Point(x*d, y*d);

	}

	Point operator /(double d) const{

		return Point(x / d, y / d);

	}

	double det(const Point &b) const{

		return x*b.y - y*b.x;

	}

	double dot(const Point &b) const{

		return x*b.x + y*b.y;

	}

};



#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))

#define crossOp(p1,p2,p3) (dcmp(cross(p1,p2,p3)))



Point isSS(Point p1, Point p2, Point q1, Point q2){

	double a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);

	return (p1*a2 + p2*a1) / (a1 + a2);

}



struct Border

{

	Point p1, p2;

	double alpha;

	void setAlpha(){

		alpha = atan2(p2.y - p1.y, p2.x - p1.x);

	}

	bool operator < ( const Border &b) const{

		int c = dcmp(alpha - b.alpha);

		if (c != 0) return c == 1;

		return crossOp(b.p1, b.p2, p1) >= 0;

	}

	bool operator == (const Border &b) const{

		return dcmp(alpha - b.alpha) == 0;

	}

};





Point isBorder(const Border &a, const Border &b){

	return isSS(a.p1, a.p2, b.p1, b.p2);

}



Border border[maxn];

Border que[maxn];

int qh, qt;

// check函数判断的是新加的半平面和由a,b两个半平面产生的交点的方向，若在半平面的左侧返回True

bool check(const Border &a, const Border &b,const Border &me){

	Point is = isBorder(a, b);

	return crossOp(me.p1, me.p2, is) > 0;

}



void convexIntersection()

{

	qh = qt = 0;

	sort(border, border + n);

	n = unique(border, border + n) - border;

	for (int i = 0; i < n; i++){

		Border cur = border[i];

		while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], cur)) --qt;

		while (qh + 1 < qt&&!check(que[qh], que[qh + 1], cur)) ++qh;

		que[qt++]=cur;

	}

	while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], que[qh])) --qt;

	while (qh + 1 < qt&&!check(que[qh], que[qh + 1], que[qt-1])) ++qh;

}



Point ps[maxn];



int main()

{

	while (cin >> n)

	{

		for (int i = 0; i < n; i++){

			scanf("%lf%lf%lf%lf", &border[i].p1.x, &border[i].p1.y, &border[i].p2.x, &border[i].p2.y);

		}

		border[n].p1.x = 0; border[n].p1.y = 0; border[n].p2.x = 10000; border[n].p2.y = 0;

		border[n+1].p1.x = 10000; border[n+1].p1.y = 0; border[n+1].p2.x = 10000; border[n+1].p2.y = 10000;

		border[n+2].p1.x = 10000; border[n+2].p1.y = 10000; border[n+2].p2.x = 0; border[n+2].p2.y = 10000;

		border[n+3].p1.x = 0; border[n+3].p1.y = 10000; border[n+3].p2.x = 0; border[n+3].p2.y = 0;

		n = n + 4;

		for (int i = 0; i < n; i++){

			border[i].setAlpha();

		}

		convexIntersection();

		int cnt = 0;

		if (qt - qh <= 2){

			puts("0.0"); continue;

		}

		for (int i = qh; i < qt; i++){

			int nxt = i + 1 == qt ? qh : i + 1;

			ps[cnt++] = isBorder(que[i], que[nxt]);

		}

		double area = 0;

		for (int i = 0; i < cnt; i++){

			area += ps[i].det(ps[(i + 1) % cnt]);

		}

		area /= 2;

		area = fabs(area);

		printf("%.1lf\n", area);

	}

	return 0;

}