【CQOI2006】凸多边形

题面

题解

半平面交模板题

今年(mathrm{PKUWC;D2T3})其实只要稍微会一些计算几何的知识就至少有(76pts)了

于是准备开始恶补计算几何

其实这道题我从去PKUWC前开始做，做到现在

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));

const double INF(1000.);
const int maxn(10010);
struct point { double x, y; };
struct line { point x, y; };
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 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; }

point S[maxn]; int top;
point Intersection(line a, line b)
{
	point c = b.x - a.x;
	double t = cross(b.y, c) / cross(b.y, a.y);
	return a.x + a.y * t;
}

inline void prepare()
{
	top = 0;
	S[++top] = (point) {INF,  INF};
	S[++top] = (point) {-INF, INF};
	S[++top] = (point) {-INF, -INF};
	S[++top] = (point) {INF, -INF};
}

void add_line(line a)
{
	static point tmp[maxn]; int tot = 0;
	S[top + 1] = S[1];
	for(RG int i = 1; i <= top; i++)
	{
		double p = cross(a.y, S[i] - a.x);
		double q = cross(a.y, S[i + 1] - a.x);
		if(p >= 0) tmp[++tot] = S[i];
		if(p * q < 0) tmp[++tot] = Intersection(a, (line) {S[i], S[i + 1] - S[i]});
	}
	top = tot;
	for(RG int i = 1; i <= top; i++) S[i] = tmp[i];
}

point p[maxn];
int main()
{
	int T, n; scanf("%d", &T);
	prepare();
	while(T--)
	{
		scanf("%d", &n);
		for(RG int i = 1; i <= n; i++)
			scanf("%lf%lf", &p[i].x, &p[i].y);
		p[n + 1] = p[1];
		for(RG int i = 1; i <= n; i++)
			add_line((line) {p[i], p[i + 1] - p[i]});
	}
	double ans = 0; S[top + 1] = S[1];
	for(RG int i = 2; i <= top; i++) ans += cross(S[i] - S[1], S[i + 1] - S[1]);
	printf("%.3lf
", ans / 2.);
	return 0;
}