【BZOJ 1069】 [SCOI2007]最大土地面积

1069: [SCOI2007]最大土地面积Time Limit:1 SecMemory Limit:162 MBSubmit:1677Solved:588[Submit][Status]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

50 01 01 10 10.5 0.5

Sample Output

1.000

HINT

数据范围 n<=2000, |x|,|y|<=100000

旋转卡壳求最大四边形面积~

四边形就是两个三角形~所以做完【POJ 2079】这道题就轻而易举了，只要再维护一下线段另一端的对踵点即可。

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#define eps 1e-9using namespace std;struct Point{double x,y;}a[50005];int n,tail,h[50005];bool cmp(Point a,Point b){if (a.x==b.x) return a.y<b.y;return a.x<b.x;}double Cross(Point a,Point b,Point c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}void Graham(){sort(a+1,a+1+n,cmp);tail=0;for (int i=1;i<=n;i++){while (tail>=2&&Cross(a[h[tail-2]],a[h[tail-1]],a[i])<eps)tail–;h[tail++]=i;}int tmp=tail;h[tail++]=n-1;for (int i=n-2;i;i–){while (tail>tmp&&Cross(a[h[tail-2]],a[h[tail-1]],a[i])<eps)tail–;h[tail++]=i;}tail–;}double Rotating_calipers(){double ans=0.0;for (int i=0;i<tail-2;i++){int x=i+1,y=i+3;for (int j=i+2;j<tail;j++){while (fabs(Cross(a[h[i]],a[h[j]],a[h[x+1]]))>fabs(Cross(a[h[i]],a[h[j]],a[h[x]])))x=(x+1)%tail;while (fabs(Cross(a[h[i]],a[h[j]],a[h[y+1]]))>fabs(Cross(a[h[i]],a[h[j]],a[h[y]])))y=(y+1)%tail;ans=max(ans,fabs(Cross(a[h[i]],a[h[j]],a[h[x]]))+fabs(Cross(a[h[i]],a[h[j]],a[h[y]])));}}return ans/2.0000;}int main(){while (scanf("%d",&n)!=EOF){for (int i=1;i<=n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);Graham();printf("%.3lf\n",Rotating_calipers());}return 0;}

把艰辛的劳作看作是生命的必然，