poj1113凸包

就是求凸包的周长加以l为半径的圆周长，证明略

由于之前写过叉积，所以graham扫描算法不是很难理解

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=1000+10,maxn=500+100,inf=0x3f3f3f;

struct point{
    double x,y;
};
point p[N],s[N];
int top,n;
double dir(point p1,point p2,point p3)
{
    return (p3.x-p1.x)*(p2.y-p1.y)-(p3.y-p1.y)*(p2.x-p1.x);
}
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool comp(point a,point b)
{
    int te=dir(p[0],a,b);
    if(te<0)return 1;
    if(te==0&&dis(p[0],a)<dis(p[0],b))return 1;
    return 0;
}
void graham()
{
    int pos,minx,miny;
    minx=miny=inf;
    for(int i=0;i<n;i++)
    {
        if(p[i].x<minx||(p[i].x==minx&&p[i].y<miny))
        {
            minx=p[i].x;
            miny=p[i].y;
            pos=i;
        }
    }
    swap(p[0],p[pos]);
    sort(p+1,p+n,comp);
    p[n]=p[0];
    s[0]=p[0],s[1]=p[1],s[2]=p[2];
    top=2;
    for(int i=3;i<=n;i++)
    {
        while(dir(s[top-1],s[top],p[i])>=0&&top>=2)top--;
        s[++top]=p[i];
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int l;
    cin>>n>>l;
    for(int i=0;i<n;i++)cin>>p[i].x>>p[i].y;
    graham();
    double ans=2*pi*l;
    for(int i=0;i<top;i++)
    {
     //   cout<<s[i].x<<" "<<s[i].y<<endl;
        if(i==top-1)ans+=dis(s[i],s[0]);
        else ans+=dis(s[i],s[i+1]);
    }
    cout<<(int)(ans+0.5)<<endl;
    return 0;
}
/*
4 3
0 0
1 0
0 1
1 1
*/

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