【POJ 1279】Art Gallery

http://poj.org/problem?id=1279

裸的半平面交的模板，按极角排序后维护一个双端队列，不要忘了最后要去除冗余，即最后一条边(或者更多的边)一定在双端队列里，但它不一定构成半平面，所以要特判。

还有平行的边也要特判，因为平行的边的交点不可求！

最后在poj上用G++交WA了好几次，换用C++就AC了/(ㄒoㄒ)/~~

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const double Pi = acos(-1.0);
const int N = 1503;
int in() {
	int k = 0, fh = 1; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar())
		if (c == '-') fh = -1;
	for(; c >= '0' && c <= '9'; c = getchar())
		k = (k << 3) + (k << 1) + c - '0';
	return k * fh;
}
 
struct Point {
	double x, y;
	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
} t[N], p[N];
struct Line {
	Point p, v;
	double ang;
	Line(Point _p = Point(0, 0), Point _v = Point(0, 0), double _ang = 0) : p(_p), v(_v), ang(_ang) {}
	bool operator < (const Line &x) const {
		return ang < x.ang;
	}
} l[N], q[N];
 
Point operator + (Point a, Point b) {return Point(a.x + b.x, a.y + b.y);}
Point operator - (Point a, Point b) {return Point(a.x - b.x, a.y - b.y);}
Point operator * (Point a, double x) {return Point(a.x * x, a.y * x);}
Point operator / (Point a, double x) {return Point(a.x / x, a.y / x);}
 
double dcmp(double x) {return fabs(x) < 1e-8 ? 0 : (x < 0 ? -1 : 1);}
double Dot(Point a, Point b) {return a.x * b.x + a.y * b.y;}
double Cross(Point a, Point b) {return a.x * b.y - a.y * b.x;}
double sqr(double x) {return x * x;}
double dis(Point a, Point b) {return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
bool onleft(Point a, Line b) {return dcmp(Cross(a - b.p, b.v)) < 0;}
Point intersection(Line a, Line b) {
	Point u; double t;
	u = a.p - b.p;
	t = Cross(b.v, u) / Cross(a.v, b.v);
	return a.p + a.v * t;
}
 
int n;
 
double mkhalf() {
	q[1] = l[1]; q[2] = l[2];
	p[1] = intersection(q[1], q[2]);
 
	int head = 1, tail = 2;
	for(int i = 3; i <= n; ++i) {
		while (head < tail && !onleft(p[tail - 1], l[i])) --tail;
		while (head < tail && !onleft(p[head], l[i])) ++head;
		q[++tail] = l[i];
		if (dcmp(Cross(q[tail].v, q[tail - 1].v)) == 0) {
			--tail;
			if (onleft(l[i].p, q[tail])) q[tail] = l[i];
		}
		if (head < tail) p[tail - 1] = intersection(q[tail - 1], q[tail]);
	}
	while (head < tail && !onleft(p[tail - 1], q[head])) --tail;
	p[tail] = intersection(q[head], q[tail]);
 
	if (tail - head <= 1) return 0;
	else {
		double ret = 0;
		p[tail + 1] = p[head];
		for(int i = head; i <= tail; ++i)
			ret += Cross(p[i], p[i + 1]);
		return ret / 2;
	}
}
 
int main() {
	int T = in();
	while (T--) {
		n = in();
		for(int i = 1; i <= n; ++i) scanf("%lf%lf", &t[i].x, &t[i].y);
		t[n + 1] = t[1];
		for(int i = 1; i <= n; ++i)
			l[i] = Line(t[i + 1], t[i] - t[i + 1], atan2(t[i].y - t[i + 1].y, t[i].x - t[i + 1].x));
 
		sort(l + 1, l + n + 1);
 
		printf("%.2lf\n", mkhalf());
	}
 
	return 0;
}

_(:з」∠)_