●POJ 3348 Cows

题链：

http://poj.org/problem?id=3348

题解：

计算几何，凸包，多边形面积

好吧，就是个裸题，没什么可讲的。

代码：

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 10050
using namespace std;
const double eps=1e-8;
int sign(double x){
	if(fabs(x)<=eps) return 0;
	return x<0?-1:1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0):x(_x),y(_y){}
	void Read(){scanf("%lf%lf",&x,&y);}
};
typedef Point Vector;
bool operator < (Point A,Point B){return sign(A.x-B.x)<0||(sign(A.x-B.x)==0&&sign(A.y-B.y)<0);}
bool operator == (Point A,Point B){return sign(A.x-B.x)==0&&sign(A.y-B.y)==0;}
Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}
double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}
Point D[MAXN],H[MAXN];
int Andrew(int dnt){
	int hnt=0,k=0;
	sort(D+1,D+dnt+1);
	dnt=unique(D+1,D+dnt+1)-D-1;
	for(int i=1;i<=dnt;i++){
		while(hnt>1&&sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1]))<=0) hnt--;
		H[++hnt]=D[i];
	} k=hnt;
	for(int i=dnt-1;i>=1;i--){
		while(hnt>k&&sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1]))<=0) hnt--;
		H[++hnt]=D[i];
	}
	return hnt;
}
double GCPA(int hnt){//Get_Convex_Polygon_Area
	double S=0;
	for(int i=1;i<hnt;i++) S+=(H[i]^H[i+1])/2;
	return S;
}
int main(){
	int N; scanf("%d",&N);
	for(int i=1;i<=N;i++) D[i].Read();
	printf("%d",(int)(GCPA(Andrew(N))/50));
	return 0;
}

原文地址：https://www.cnblogs.com/zj75211/p/8227651.html