BZOJ1069 [SCOI2007]最大土地面积

终于做出来了2333~~~~

凸包+旋转卡（qia）壳。

n = 2000，所以可以先枚举两个点作为对角线上的点。

然后由于决策单调性，另外的两个点可以o(1)求出，所以就做好了额。

计算几何太烦太烦>.<

/**************************************************************
    Problem: 1069
    User: rausen
    Language: C++
    Result: Accepted
    Time:148 ms
    Memory:872 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
#define points P
using namespace std;
typedef double lf;
const int N = 2005;
struct points{
    lf x, y;
}p[N], s[N];
int n, top;
 
 
inline lf operator * (const P a, const P b){
    return a.x * b.y - a.y * b.x;
}
 
inline P operator - (const P a, const P b){
    P tmp;
    tmp.x = a.x - b.x, tmp.y = a.y - b.y;
    return tmp;
}
 
inline lf Sqr(const lf x){
    return x * x;
}
 
inline lf dis(const P a, const P b){
    return Sqr(a.x - b.x) + Sqr(a.y - b.y);
}
 
inline bool operator < (const P a, const P b){
    lf tmp = (a - p[1]) * (b - p[1]);
    return tmp == 0 ? dis(p[1], a) < dis(p[1], b) : tmp > 0;
}
 
inline lf Area(const int x, const int y, const int z){
    return (s[z] - s[x]) * (s[y] - s[x]);
}
 
int work(){
    int k = 1;
    for (int i = 2; i <= n; ++i)
        if (p[i].y == p[k].y ? p[i].x < p[k].x : p[i].y < p[k].y) k = i;
    swap(p[1], p[k]);
    sort(p + 2, p + n + 1);
    top = 2, s[1] = p[1], s[2] = p[2];
    for (int i = 3; i <= n; ++i){
        while ((s[top] - s[top - 1]) * (p[i] - s[top]) <= 0) --top;
        s[++top] = p[i];
    }
}
 
double Calc(){
    s[top + 1] = p[1];
    lf res = 0;
    int a, b, x, y;
    for (x = 1; x <= top; ++x){
        a = x % top + 1, b = (x + 2) % top + 1;
        for (y = x + 2; y <= top; ++y){
            while (a % top + 1 != y && Area(x, y, a + 1) > Area(x, y, a))
                a = a % top + 1;
            while (b % top + 1 != x && Area(x, b + 1, y) > Area(x, b, y))
                b = b % top + 1;
            res = max(Area(x, y, a) + Area(x, b, y), res);
        }
    }
    return res;
}
 
int main(){
    scanf("%d
", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%lf%lf", &p[i].x, &p[i].y);
    work();
    printf("%.3lf
", Calc() / 2);
    return 0;
}