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POJ 3348 Cows [凸包 面积]
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 9022 | Accepted: 3992 |
Description
Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.
However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.
Input
The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and yseparated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).
Output
You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.
Sample Input
40 00 10175 075 101
Sample Output
151
Source
标准模板题
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>using namespace std;typedef long long ll;const int N=1e4+5;const double eps=1e-8;const double pi=acos(-1);inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f;}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 x<a.x||(x==a.x&&y<a.y); 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;}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 Area(Point p[],int n){ double s=0; for(int i=2;i<n;i++) s+=Cross(p[i]-p[1],p[i+1]-p[1]); return s/2;}int n,x,y;double ans;Point p[N],ch[N];int main(int argc, const char * argv[]) { n=read(); for(int i=1;i<=n;i++) p[i].x=read(),p[i].y=read(); int m=ConvexHull(p,n,ch); printf("%d",int(Area(ch,m))/50); return 0;}
POJ 3348 Cows [凸包 面积]