先求凸包，然后再用旋轉(zhuǎn)卡殼方法求解。
具體做法是枚舉三角形的第一個點i，設j = i + 1，k = j + 1。然后做以下操作：
1.計算i，j，k構(gòu)成的三角形面積a1和i，j，k + 1構(gòu)成的三角形面積a2，如果a2 < a1，則進行下一步，否則k++，重復此步。
2.記錄此時的三角形面積b，如果b < preb（就是上一個j對應的三角形面積）j++，轉(zhuǎn)第一步，否則退出。
可以證明這個算法的復雜度為O(n
2)。具體實現(xiàn)見代碼。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-9-12 19:48:12
File Name: b.cpp
Description:
************************************************************************/
#include <iostream>
using namespace std;
#define out(x) (cout<<#x<<": "<<x<<endl)
const int maxint=0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template<class T>void show(T a, int n)
{for(int i=0; i<n; ++i) cout<<a[i]<<' '; cout<<endl;}
template<class T>void show(T a, int r, int l){for(int i=0; i<r; ++i)show(a[i],l);cout<<endl;}
const int maxn = 50010;
typedef struct point_t
{
int x, y;
};
typedef struct polygon_t
{
int n;
point_t p[maxn];
};
int operator <(const point_t &a, const point_t &b)
{
return a.y < b.y || a.y == b.y && a.x < b.x;
}
point_t operator -(const point_t &a, const point_t &b)
{
point_t ret;
ret.x = a.x - b.x;
ret.y = a.y - b.y;
return ret;
}
inline int cross(const point_t &a, const point_t &b)
{
return a.x * b.y - a.y * b.x;
}
int turn_left(const point_t &a, const point_t &b, const point_t &c)
{
return cross(b - a, c - b) > 0;
}
class point_set_c
{
public:
void init(int _n, point_t _p[]);
polygon_t convex_hull();
private:
int n;
point_t p[maxn];
};
void point_set_c::init(int _n, point_t _p[maxn])
{
n = _n;
for (int i = 0; i < n; i++)
p[i] = _p[i];
}
polygon_t point_set_c::convex_hull()
{
int stack[maxn];
int top = 1;
stack[0] = 0;
sort(p, p + n);
for (int i = 1; i < n;)
{
if (top == 1 || turn_left(p[stack[top - 2]], p[stack[top - 1]], p[i]))
stack[top++] = i++;
else top--;
}
int t_top = top;
for (int i = n - 2; i >= 0;)
{
if (top == t_top || turn_left(p[stack[top - 2]], p[stack[top - 1]], p[i]))
stack[top++] = i--;
else top--;
}
polygon_t ret;
ret.n = 0;
for (int i = 0; i < top - 1; i++)
ret.p[ret.n++] = p[stack[i]];
return ret;
}
point_t p[maxn];
int n;
point_set_c ps;
polygon_t poly;
int main()
{
while (scanf("%d", &n), n != -1)
{
for (int i = 0; i < n; i++)
scanf("%d%d", &p[i].x, &p[i].y);
ps.init(n, p);
poly = ps.convex_hull();
int ans = 0;
for (int i = 0; i < poly.n; i++)
{
int j = i + 1, k = j + 1;
if (k >= poly.n)
break;
int area;
int pre_area = -1;
while (j < poly.n - 1)
{
area = cross(poly.p[j] - poly.p[i], poly.p[k] - poly.p[i]);
while (k < poly.n - 1)
{
int tmp = cross(poly.p[j] - poly.p[i], poly.p[k + 1] - poly.p[i]);
if (tmp > area)
{
k++;
area = tmp;
}
else
break;
}
if (area < pre_area)
break;
pre_area = area;
ans >?= area;
j++;
if (j >= k)
{
k = j + 1;
if (k >= poly.n)
break;
}
}
}
printf("%.2lf\n", ans / 2.0);
}
return 0;
}