poj1113Wall 求凸包周长 Graham扫描法

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef pair<int ,int > ll;
ll num,dot[1010];
int i;
const double pi=3.1415926535898;
ll operator -(ll a,ll b)
{
return make_pair(a.first-b.first,a.second-b.second);
}
bool cmp(ll a,ll b)
{
return (a.first!=b.first?a.first<b.first:a.second<b.second);
}
int cross(ll a,ll b)
{
return a.first*b.second-b.first*a.second;
}
int d2(ll a,ll b)
{
return (a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
}
double d1(ll a,ll b)
{
return sqrt((double)d2(a,b));
}
bool cmp1(ll a,ll b)
{
int r=cross(a-dot[0],b-dot[0]);
return (r!=0?r>0:d2(a,dot[0])<d2(b,dot[0]));
}
void in()
{
cin>>num.first>>num.second;
for(i=0;i<num.first;i++)
cin>>dot[i].first>>dot[i].second;
}
void work()
{
sort(dot,dot+num.first,cmp);
sort(dot,dot+num.first,cmp1);

ll box[1010]={dot[0],dot[1]};
int tail=1;
for(i=2;i<num.first;)
{
if(cross(box[tail]-box[tail-1],dot[i]-box[tail-1])>=0)
box[++tail]=dot[i++];
else
tail--;
}

double ans=d1(box[0],box[tail])+2*pi*num.second;
for(i=0;i<tail;i++)
ans+=d1(box[i],box[i+1]);
printf("%.0lf\n",ans);
}
int main()
{
in();
work();
}