poj 1113 Wall

题目链接：http://poj.org/problem?id=1113

题目大意：给出点集和一个长度L，要求用最短长度的围墙把所有点集围住，并且围墙每一处距离所有点的距离最少为L，求围墙的长度。

解法：凸包+以L为半径的圆的周长。以题目中的图为例，两点之间的围墙长度之和正好就是凸包的长度，再加上每个点的拐角处（注意此处为弧，才能保证城墙距离点的距离最短）的长度。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
#define exp 1e-10
#define PI 3.141592654
using namespace std;
int n,L;
struct Point
{
    double x,y;
    Point (double x=0,double y=0):x(x),y(y){}
    friend bool operator < (Point a,Point b)
    {
        if (a.x!=b.x) return a.x<b.x;
        return a.y<b.y;
    }
}an[1000+10],bn[1000+10];
typedef Point Vector;
Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
int dcmp(double x)
{
    if (fabs(x)<exp) return 0;
    else return x<0 ? -1 : 1;
}
double cross(Vector A,Vector B)
{
    return A.x*B.y-B.x*A.y;
}
int ConvexHull(Point *p,int n,Point *ch)
{
    sort(p,p+n);
    int m=0;
    for (int i=0 ;i<n ;i++)
    {
        while (m>1 && dcmp(cross(ch[m-1]-ch[m-2],p[i]-ch[m-1]))<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    for (int i=n-2 ;i>=0 ;i--)
    {
        while (k>m && dcmp(cross(ch[k-1]-ch[k-2],p[i]-ch[k-1]))<=0) k--;
        ch[k++]=p[i];
    }
    ch[k++]=ch[0];
    if (n>1) k--;
    return k;
}
int main()
{
    while (cin>>n>>L)
    {
        for (int i=0 ;i<n ;i++)
        scanf("%lf%lf",&an[i].x,&an[i].y);

        int k=ConvexHull(an,n,bn);
        double sum=0;
        sum += 2*PI*L;
        for (int i=0 ;i<k ;i++)
        {
            sum += sqrt((bn[i].x-bn[i+1].x)*(bn[i].x-bn[i+1].x)+(bn[i].y-bn[i+1].y)*(bn[i].y-bn[i+1].y));
        }
        printf("%.0f
",sum);
    }
    return 0;
}