BZOJ 1069: [SCOI2007]最大土地面积 [旋转卡壳]

1069: [SCOI2007]最大土地面积

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2978  Solved: 1173
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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

---

4边形呵呵

枚举对角线，就是两个三角形啊....并且还是两个点确定的...卡就行了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=;
const double eps=1e-;

inline int sgn(double x){
    if(abs(x)<eps) return ;
    else return x<?-:;
}

struct Vector{
    double x,y;
    Vector(double a=,double b=):x(a),y(b){}
    bool operator <(const Vector &a)const{
        return sgn(x-a.x)<||(sgn(x-a.x)==&&sgn(y-a.y)<);
    }
};
typedef Vector Point;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==&&sgn(a.y-b.y)==;}
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}

double Len(Vector a){return sqrt(Dot(a,a));}
double Len2(Vector a){return Dot(a,a);}
double DisTL(Point p,Point a,Point b){
    Vector v1=p-a,v2=b-a;
    return abs(Cross(v1,v2)/Len(v2));
}
int ConvexHull(Point p[],int n,Point ch[]){
    sort(p+,p++n);
    int m=;
    for(int i=;i<=n;i++){
        while(m>&&sgn(Cross(ch[m]-ch[m-],p[i]-ch[m-]))<=) m--;
        ch[++m]=p[i];
    }
    int k=m;
    for(int i=n-;i>=;i--){
        while(m>k&&sgn(Cross(ch[m]-ch[m-],p[i]-ch[m-]))<=) m--;
        ch[++m]=p[i];
    }
    if(n>) m--;
    return m;
}
double S(Vector a,Vector b){return abs(Cross(a,b));}
double RotatingCalipers(Point p[],int n){
    if(n<) return ;
    if(n==) return abs(Cross(p[]-p[],p[]-p[]))/+abs(Cross(p[]-p[],p[]-p[]))/;
    double ans=;
    p[n+]=p[];
    for(int i=;i<=n-;i++){
        int k=i+,l=(i+)%n+;
        for(int j=i+;j<=n;j++){
            while(k+<j&&sgn(S(p[k]-p[i],p[j]-p[i])-S(p[k+]-p[i],p[j]-p[i]))<) k=k+;
            while(l%n+!=i&&sgn(S(p[l]-p[i],p[j]-p[i])-S(p[l%n+]-p[i],p[j]-p[i]))<) l=l%n+;
            ans=max(ans,S(p[k]-p[i],p[j]-p[i])/+S(p[l]-p[i],p[j]-p[i])/);
        }
    }
    return ans;
}

int n;
Point p[N],ch[N];
int main(int argc, const char * argv[]) {
    scanf("%d",&n);
    for(int i=;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
    n=ConvexHull(p,n,ch);
    double ans=RotatingCalipers(ch,n);
    printf("%.3f",ans);
}