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[POJ1113]Wall(凸包)

2017-01-03 08:37 337 查看

题目描述

传送门

题意:给出一个多边形,要在外围距离r围一圈围墙,求围墙的长度

题解

nπ−(n−3)π=2π

所以就是求凸包周长然后加上一个圆周的长

从图里很清楚能看出来

Graham求凸包裸题

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 1005

const double pi=acos(-1.0);
const double eps=1e-9;
int dcmp(double x)
{
if (x<=eps&&x>=-eps) return 0;
return (x>0)?1:-1;
}
struct Vector
{
double x,y;
Vector(double X=0,double Y=0)
{
x=X,y=Y;
}
bool operator < (const Vector a) const
{
return x<a.x||(x==a.x&&y<a.y);
}
};
typedef Vector Point;
Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x,A.y-B.y);}

int n,top;
double r,x,y,ans;
Point p
,stack
;

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;
}
double Len(Vector A)
{
return sqrt(Dot(A,A));
}
void Graham()
{
sort(p+1,p+n+1);
for (int i=1;i<=n;++i)
{
while (top>1&&dcmp(Cross(stack[top]-stack[top-1],p[i]-stack[top-1]))<=0)
--top;
stack[++top]=p[i];
}
int k=top;
for (int i=n-1;i>=1;--i)
{
while (top>k&&dcmp(Cross(stack[top]-stack[top-1],p[i]-stack[top-1]))<=0)
--top;
stack[++top]=p[i];
}
if (n>1) --top;
}
int main()
{
freopen("input.in","r",stdin);
scanf("%d%lf",&n,&r);
for (int i=1;i<=n;++i)
{
scanf("%lf%lf",&x,&y);
p[i]=Point(x,y);
}
Graham();
for (int i=1;i<=top;++i)
ans+=Len(stack[i]-stack[i%top+1]);
ans+=2.0*pi*r;
printf("%.0lf\n",ans);
}
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