【SCOI2015】小凸想跑步

题面

题解

推波柿子：

设点(A(x_a, y_a), B(x_b, y_b), C(x_c, y_c), D(x_d, y_d), P(x, y))

(vec{a} = (x_b - x_a, y_b - y_a), vec{b} = (x_d - x_c, y_d - y_c))

(overrightarrow{AP} = (x - x_a, y - y_a), overrightarrow{CP} = (x - x_c, y - y_c))

(vec{a} imes overrightarrow{AP} = (x_b - x_a)(y - y_a) - (x_b - y_a)(x - x_a))

(vec{b} imes overrightarrow{CP} = (x_d - x_c)(y - y_c) - (y_d - y_c)(x - x_c))

由题目，(vec{a} imes overrightarrow{AP} < vec{b} imes overrightarrow{CP})，那么有

[egin{aligned} &(x_b - x_a)(y - y_a) - (x_b - y_a)(x - x_a) < (x_d - x_c)(y - y_c) - (y_d - y_c)(x - x_c) \ Rightarrow & (x_b - x_a + x_d - x_c)y - (y_b - y_a - y_d + y_c)x + (y_b x_a - x_b y_a + y_d x_c - x_d y_c) < 0 end{aligned} ]

然后这个就是一个裸的半平面交了。

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(200010);
struct point { double x, y; } p[maxn];
struct line { point x, y; double ang; } L[maxn];
typedef point vector;

inline vector operator + (const vector &lhs, const vector &rhs)
	{ return (vector) {lhs.x + rhs.x, lhs.y + rhs.y}; }
inline vector operator - (const vector &lhs, const vector &rhs)
	{ return (vector) {lhs.x - rhs.x, lhs.y - rhs.y}; }
inline vector operator * (const vector &lhs, const double &rhs)
	{ return (vector) {lhs.x * rhs,   lhs.y * rhs};   }
inline double operator * (const vector &lhs, const vector &rhs)
	{ return lhs.x * rhs.x + lhs.y * rhs.y; }
inline double cross(const vector &lhs, const vector &rhs)
	{ return lhs.x * rhs.y - lhs.y * rhs.x; }
inline line make_line(const point &x, const point &y)
	{ return (line) {x, y, atan2(y.y, y.x)}; }
inline bool operator < (const line &lhs, const line &rhs)
	{ return lhs.ang < rhs.ang; }
inline bool isLeft(const point &a, const line &b)
	{ return cross(b.y, (a - b.x)) > 0; }
point Intersection(const line &a, const line &b)
	{ return a.x + a.y * (cross(b.y, b.x - a.x) / cross(b.y, a.y)); }

int n, m;
double Area, ans;

void HalfPlane()
{
	int l, r = 1; std::sort(L + 1, L + m + 1);
	for(RG int i = 2; i <= m; i++)
		if(L[i].ang != L[r].ang) L[++r] = L[i];
		else if(isLeft(L[i].x, L[r])) L[r] = L[i];
	m = r, l = r = 1;
	for(RG int i = 2; i <= m; i++)
	{
		while(l < r && !isLeft(p[r], L[i])) --r;
		while(l < r && !isLeft(p[l + 1], L[i])) ++l;
		L[++r] = L[i];
		if(l < r) p[r] = Intersection(L[r], L[r - 1]);
	}
	while(l < r && !isLeft(p[r], L[l])) --r;
	p[l] = p[r + 1] = Intersection(L[r], L[l]);
	for(RG int i = l; i <= r; i++) ans += cross(p[i], p[i + 1]);
	ans /= Area;
}

int main()
{
	n = read();
	for(RG int i = 0; i < n; i++)
		p[i] = (point) {(double)read(), (double)read()};
	p[n] = p[0];
	for(RG int i = 0; i < n; i++)
		L[++m] = make_line(p[i], p[i + 1] - p[i]),
		Area += cross(p[i], p[i + 1]);
	for(RG int i = 1; i < n; i++)
	{
		double a = p[1].x - p[0].x + p[i].x - p[i + 1].x;
		double b = p[1].y - p[0].y + p[i].y - p[i + 1].y;
		double c = p[1].x * p[0].y + p[i].x * p[i + 1].y
			- p[0].x * p[1].y - p[i + 1].x * p[i].y;
		if(a) L[++m] = make_line((point) {0, c / a}, (point) {-a, -b});
		else if(b) L[++m] = make_line((point) {-c / b, 0}, (point) {0, -b});
	}
	HalfPlane();
	printf("%.4lf
", ans);
	return 0;
}