[USACO5.1]二维凸包模板

传送门

Description

(n)个点凸包的周长

Solution 

计算几何打暴力必备


Code 

#include<bits/stdc++.h>
#define reg register
#define ll long long
#define db double
using namespace std;

const int MN=1e5+5;
const db eps=1e-8;
int n;
struct Point{
	db x,y;
	Point(db x=0,db y=0):x(x),y(y){}
	db mo(){return sqrt(1.*x*x+1.*y*y);}
	Point operator-(const Point&o)const{return Point(x-o.x,y-o.y);}
	db operator^(const Point&o)const{return x*o.y-y*o.x;}
	bool operator <(const Point&o)const{return x<o.x||(x==o.x&&y<o.y);}
}p[MN];
int u[MN],d[MN],tu,td;

int main()
{
	scanf("%d",&n);
	reg int i;
	for(i=1;i<=n;++i) scanf("%lf%lf",&p[i].x,&p[i].y);
	std::sort(p+1,p+n+1);
	for(i=1;i<=n;++i)
	{
		while(tu>1&&((p[u[tu]]-p[u[tu-1]])^(p[i]-p[u[tu-1]]))>-eps) --tu;
		u[++tu]=i;
	}
	for(i=n;i;--i)
	{
		while(td>1&&((p[d[td]]-p[d[td-1]])^(p[i]-p[d[td-1]]))>-eps) --td;
		d[++td]=i;
	}
	db ans=0.;
	for(i=1;i<tu;++i) ans+=(p[u[i+1]]-p[u[i]]).mo();
	for(i=1;i<td;++i) ans+=(p[d[i+1]]-p[d[i]]).mo();
	return 0*printf("%.2lf
",ans); 
}


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原文地址:https://www.cnblogs.com/PaperCloud/p/10920188.html