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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\n", ++c, ans);	}    return 0;}

LightOJ 1239 - Convex Fence 凸包周长