【BZOJ 1069】【SCOI 2007】最大土地面积 凸包+旋转卡壳

因为凸壳上对踵点的单调性所以旋转卡壳线性绕一圈就可以啦啦啦～～～

先求凸包，然后旋转卡壳记录$sum1$和$sum2$，最后统计答案就可以了

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define read(x) x=getint()
#define N 2003
using namespace std;
inline int dcmp(double x) {return fabs(x) < 1e-6 ? 0 : (x < 0 ? -1 : 1);}
struct Point {
	double x, y;
	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
} a[N], tu[N];
Point operator - (Point a, Point b) {
	return Point(a.x - b.x, a.y - b.y);
}
inline double Cross(Point a, Point b) {
	return a.x * b.y - a.y * b.x;
}
inline double S(Point a, Point b, Point c) {
	return Cross(a - c, b - c);
}

int n, top = 0;
double sum1[N][N], sum2[N][N];
inline bool cmp(Point X, Point Y) {
	return X.y == Y.y ? X.x < Y.x : X.y < Y.y;
}
inline void mktb() {
	for(int i = 1; i <= n; ++i) {
		while (top > 1 && dcmp(S(tu[top], a[i], tu[top-1])) != 1)
			--top;
		tu[++top] = a[i];
	}
	int k = top;
	for(int i = n-1; i > 0; --i) {
		while (top > k && dcmp(S(tu[top], a[i], tu[top - 1])) != 1)
			--top;
		tu[++top] = a[i];
	}
	tu[0] = tu[top - 1];
	n = top - 1;
}
inline void mksum() {
	int nxt, j;
	for(int i = 0; i < n; ++i) {
		nxt = (i + 2) % n;
		for(int tmp = 1; tmp <= n - 2; ++tmp) {
			j = (i + tmp) % n;
			while (S(tu[j], tu[nxt], tu[i]) < S(tu[j], tu[(nxt + 1) % n], tu[i]))
				nxt = (nxt + 1) % n;
			sum1[i][j] = S(tu[j], tu[nxt], tu[i]);
		}
		nxt = (i - 2 + n) % n;
		for(int tmp = 1; tmp <= n - 2; ++tmp) {
			j = (i - tmp + n) % n;
			while (S(tu[nxt], tu[j], tu[i]) < S(tu[(nxt - 1 + n) % n], tu[j], tu[i]))
				nxt = (nxt - 1 + n) % n;
			sum2[i][j] = S(tu[nxt], tu[j], tu[i]);
		}
	}
}
inline void AC() {
	double ans = 0;
	for(int i = 0; i < n - 2; ++i)
		for(int j = i + 2; j < n; ++j)
			ans = max(ans, sum1[i][j] + sum2[i][j]);
	printf("%.3lf
", ans / 2);
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%lf%lf", &a[i].x, &a[i].y);
	sort(a + 1, a + n + 1, cmp);
	mktb();
	mksum();
	AC();
	return 0;
}

没什么可说的了╮(๑•́ ₃•̀๑)╭