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: 2938  Solved: 1149
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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

Source

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题解：旋转卡壳

枚举对角线，在两侧求最大面积三角形。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 2003#define eps 1e-7using namespace std;struct vector{double x,y;vector (double X=0,double Y=0){x=X,y=Y;}}a[N],ch[N];int n,m,cnt;double ans;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 t){return vector (a.x*t,a.y*t);}bool operator <(vector a,vector b){return a.x<b.x||a.x==b.x&&a.y<b.y;}int dcmp(double x){if (fabs(x)<eps) return 0;return x<0?-1:1;}double cross(vector a,vector b){return a.x*b.y-a.y*b.x;}double dot(vector a,vector b){return a.x*b.x+a.y*b.y;}void convexhull(){sort(a+1,a+n+1);m=0;if (n==1){ch[m++]=a[1];return;}for (int i=1;i<=n;i++){while (m>1&&cross(ch[m-1]-ch[m-2],a[i]-ch[m-2])<=0) m--;ch[m++]=a[i];}int k=m;for (int i=n-1;i>=1;i--){while (m>k&&cross(ch[m-1]-ch[m-2],a[i]-ch[m-2])<=0) m--;ch[m++]=a[i];}m--;}void rotating(){int i,j,k,k1;for (int i=0;i<m-2;i++){ k=(i+3)%m; k1=(i+1)%m; for (int j=i+2;j<m;j++){while (fabs(cross(ch[j]-ch[i],ch[k]-ch[i]))<fabs(cross(ch[j]-ch[i],ch[(k+1)%m]-ch[i]))&&(k+1)%m!=i) k=(k+1)%m;while (fabs(cross(ch[j]-ch[i],ch[k1]-ch[i]))<fabs(cross(ch[j]-ch[i],ch[(k1+1)%m]-ch[i]))&&(k1+1)%m!=j) k1=(k1+1)%m;ans=max(ans,fabs(cross(ch[j]-ch[i],ch[k]-ch[i]))+fabs(cross(ch[j]-ch[i],ch[k1]-ch[i])));  }    }}int main(){freopen("land.in","r",stdin);freopen("land.out","w",stdout);scanf("%d",&n);for (int i=1;i<=n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);convexhull();/*if (m<4) {printf("0.000\n");return 0;}*///for (int i=0;i<m;i++) cout<<ch[i].x<<" "<<ch[i].y<<endl;ans=0;rotating();printf("%.3lf\n",ans/2.0);}