LightOJ 1239

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题意：类似POJ的宫殿围墙那道，只不过这道题数据稍微强了一点，有共线的情况

思路：求凸包周长加一个圆周长

/** @Date    : 2017-07-20 15:46:44
  * @FileName: LightOJ 1239 求凸包.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
const double Pi = acos(-1.0);
struct point
{
	double x, y;
	point(){}
	point(double _x, double _y){x = _x, y = _y;}
	point operator -(const point &b) const
	{
		return point(x - b.x, y - b.y);
	}
	double operator *(const point &b) const 
	{
		return x * b.x + y * b.y;
	}
	double operator ^(const point &b) const
	{
		return x * b.y - y * b.x;
	}
};

double xmult(point p1, point p2, point p0)  
{  
    return (p1 - p0) ^ (p2 - p0);  
}  

double distc(point a, point b)
{
	return sqrt((double)((b - a) * (b - a)));
}
int sign(double x)
{
	if(fabs(x) < eps)
		return 0;
	if(x < 0)
		return -1;
	else 
		return 1;
}

////////
int n;
point stk[N];
point p[N];

int cmpC(point a, point b)//水平序排序
{
	return sign(a.x - b.x) < 0 || (sign(a.x - b.x) == 0 && sign(a.y - b.y) < 0);
}

int Graham(point *p, int n)//水平序
{
	sort(p, p + n, cmpC);
	int top = 0;
	for(int i = 0; i < n; i++)
	{
		while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], p[i])) < 0)
			top--;
		stk[top++] = p[i];
	}
	int tmp = top;
	for(int i = n - 2; i >= 0; i--)
	{
		while(top > tmp && sign(xmult(stk[top - 2],stk[top - 1] ,p[i] )) < 0)
			top--;
		stk[top++] = p[i];
	}
	if(n > 1)
		top--;
	return top;
}

int main()
{
	int T;
	cin >> T;
	int c = 0;
	while(T--)
	{
		int n;
		double l;
		cin >> n >> l;
		double x, y;
		for(int i = 0; i < n; i++)
		{
			scanf("%lf%lf", &x, &y);
			p[i] = point(x, y);
		}
		int m = Graham(p, n);
		double ans = 2 * Pi * l;
		stk[m++] = stk[0];//注意只有直线的情况
		for(int i = 0; i < m - 1; i++)
			ans += distc(stk[i], stk[i + 1]);
		printf("Case %d: %.10lf
", ++c, ans);
	}
    return 0;
}