Cows POJ - 3348
Cows
POJ - 3348题意：给你一些树的坐标，让你来建筑面积最大的牛场，问能养多少头牛，一个牛的面积50
思路：求凸包面积即可，先求出凸包，然后再求ch[]里面的凸包顶点集所围成的多边形的面积即可
1 // 2 // Created by HJYL on 2020/2/3. 3 // 4 #include<iostream> 5 #include<cstring> 6 #include<cstdio> 7 #include<cmath> 8 #include<algorithm> 9 using namespace std; 10 const double eps=1e-8; 11 const int maxn=1e5+10; 12 const int INF=0x3ffff; 13 struct Point{ 14 double x,y; 15 Point(double x=0,double y=0):x(x),y(y){} 16 Point operator - (Point const &b)const 17 { 18 return Point(x-b.x ,y-b.y); 19 } 20 bool operator < (Point const &c)const{ 21 if(x==c.x) 22 return y<c.y; 23 return x<c.x; 24 } 25 }p[maxn],ch[maxn]; 26 int n,m; 27 double Cross(Point A,Point B) 28 { 29 return (A.x*B.y)-(A.y*B.x); 30 } 31 double PolygonArea(Point* p, int n){//p为端点集合，n为端点个数 32 double s = 0; 33 for(int i = 1; i < n-1; ++i) 34 s += Cross(p[i]-p[0], p[i+1]-p[0]); 35 return s; 36 } 37 void ConvexHull() ///**求凸包*/ 38 { 39 sort(p,p+n);//n顶点数 40 m = 0; 41 for(int i = 0; i < n; i++) 42 { 43 while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; 44 ch[m++] = p[i]; 45 } 46 int k = m; 47 for(int i = n-2; i >= 0; i--) 48 { 49 while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; 50 ch[m++] = p[i]; 51 } 52 if(n > 1) m--; 53 } 54 int main() 55 { 56 while(~scanf("%d",&n)) 57 { 58 for(int i=0;i<n;i++) 59 scanf("%lf%lf",&p[i].x,&p[i].y); 60 ConvexHull(); 61 double ss=PolygonArea(ch,m)/2.0; 62 int num=ss/50.0; 63 printf("%d\n",num); 64 } 65 return 0; 66 }
