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();
}

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