[BZOJ]1069 [SCOI2007] 最大土地面积 凸包 + 旋转卡壳

1069: [SCOI2007]最大土地面积

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 3720   Solved: 1471
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Description

　　在某块平面土地上有N个点，你可以选择其中的任意四个点，将这片土地围起来，当然，你希望这四个点围成
的多边形面积最大。

Input

　　第1行一个正整数N，接下来N行，每行2个数x,y，表示该点的横坐标和纵坐标。

Output

　　最大的多边形面积，答案精确到小数点后3位。

Sample Input

5
0 0
1 0
1 1
0 1
0.5 0.5

Sample Output

1.000

HINT

数据范围 n<=2000, |x|,|y|<=100000

Source

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　　首先很显然四个点一定在凸包上， 其次... n只有两千? 暴力枚举对角线， 以对角线为底， 分别在两边找最大面积三角形. 很容易发现这是单调的， 像旋(xuan)转(zhuan)卡(qia)壳(ke)那样搞即可.

#include<bits/stdc++.h>
#define pp pop_back
#define pb push_back
using namespace std;
int n, m;
double ans;
struct Vector {
	double x, y;
	Vector() {}
	Vector(double x, double y) : x(x), y(y) {}
	inline friend bool operator < (const Vector &r, const Vector &s) {
		return r.x < s.x || (r.x == s.x && r.y < s.y);
	}
	inline friend Vector operator - (const Vector &r, const Vector &s) {
		return Vector(r.x - s.x, r.y - s.y);
	}
};
typedef Vector Point;
vector<Point> pts, cvx;
inline double cross (const Vector &r, const Vector &s) {
	return r.x * s.y - r.y * s.x;
}
inline double area(const Point &A, const Point &B, const Point &C) {
	return cross(B - A, C - A);
}
inline void Convex_Hull() {
	sort(pts.begin(), pts.end());
	for (int i = 0; i < n; ++ i) {
		while (m > 1 && area(cvx[m - 2], cvx[m - 1], pts[i]) >= 0) cvx.pp(), m --;
		cvx.pb(pts[i]), m ++;
	}
	int k = m;
	for (int i = n - 2; ~i; -- i) {
		while (m > k && area(cvx[m - 2], cvx[m - 1], pts[i]) >= 0) cvx.pp(), m --;
		cvx.pb(pts[i]), m ++;
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++ i) {
		double x, y;
		scanf("%lf%lf", &x, &y);
		pts.pb(Point(x, y));
	}
	Convex_Hull();
	for (int i = 0; i < m - 1; ++ i) {
		int h = i + 1, t = i + 3;
		for (int j = i + 2; j < m - 2; ++ j) {
			while (h < j - 2 && area(cvx[i], cvx[j], cvx[h]) <= area(cvx[i], cvx[j], cvx[h + 1])) h ++;
			while (t <= j || (t < m - 2 && area(cvx[i], cvx[t], cvx[j]) <= area(cvx[i], cvx[t + 1], cvx[j]))) t ++;
			ans = max(ans, area(cvx[i], cvx[j], cvx[h]) + area(cvx[i], cvx[t], cvx[j]));
		}
	}
	printf("%.3f\n", ans / 2);
}