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=2005;
const double eps=1e-8;

inline int sgn(double x){
if(abs(x)<eps) return 0;
else return x<0?-1:1;
}

struct Vector{
double x,y;
Vector(double a=0,double b=0):x(a),y(b){}
bool operator <(const Vector &a)const{
return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0);
}
};
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)==0&&sgn(a.y-b.y)==0;}
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+1,p+1+n);
int m=0;
for(int i=1;i<=n;i++){
while(m>1&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--;
ch[++m]=p[i];
}
int k=m;
for(int i=n-1;i>=1;i--){
while(m>k&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--;
ch[++m]=p[i];
}
if(n>1) m--;
return m;
}
double S(Vector a,Vector b){return abs(Cross(a,b));}
double RotatingCalipers(Point p[],int n){
if(n<3) return 0;
if(n==4) return abs(Cross(p[3]-p[1],p[2]-p[1]))/2+abs(Cross(p[3]-p[1],p[4]-p[1]))/2;
double ans=0;
p[n+1]=p[1];
for(int i=1;i<=n-2;i++){
int k=i+1,l=(i+2)%n+1;
for(int j=i+2;j<=n;j++){
while(k+1<j&&sgn(S(p[k]-p[i],p[j]-p[i])-S(p[k+1]-p[i],p[j]-p[i]))<0) k=k+1;
while(l%n+1!=i&&sgn(S(p[l]-p[i],p[j]-p[i])-S(p[l%n+1]-p[i],p[j]-p[i]))<0) l=l%n+1;
ans=max(ans,S(p[k]-p[i],p[j]-p[i])/2+S(p[l]-p[i],p[j]-p[i])/2);
}
}
return ans;
}

int n;
Point p
,ch
;
int main(int argc, const char * argv[]) {
scanf("%d",&n);
for(int i=1;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);
}