poj1113Wall凸包入门题 模板

Description

Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall. 

Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements. 

The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.

Input

The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King's castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle. 

Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.

Output

Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King's requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates.

Sample Input

9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200

Sample Output

1628

据说明天有省赛 后天有网络赛 突然想到有这么一块不会的 学习一下 ~

水一题  就是求凸包的长度+圆形的周长 这么唱的代码比赛的时候怎么办

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
const double inf=1e20;
const int maxp=1111;
int dblcmp(double d)
{
    if (fabs(d)<eps)return 0;
    return d>eps?1:-1;
}
inline double sqr(double x){return x*x;}

struct point
{
    double x,y;
    point(){}
    point(double _x,double _y):
    x(_x),y(_y){};
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    void output()
    {
        printf("%.2f %.2f\n",x,y);
    }
    bool operator==(point a)const
    {
        return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
    }
    bool operator<(point a)const
    {
        return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
    }
    double len()
    {
        return hypot(x,y);
    }
    double len2()
    {
        return x*x+y*y;
    }
    double distance(point p)
    {
        return hypot(x-p.x,y-p.y);
    }
    point add(point p)
    {
        return point(x+p.x,y+p.y);
    }
    point sub(point p)
    {
        return point(x-p.x,y-p.y);
    }
    point mul(double b)
    {
        return point(x*b,y*b);
    }
    point div(double b)
    {
        return point(x/b,y/b);
    }
    double dot(point p)
    {
        return x*p.x+y*p.y;
    }
    double det(point p)
    {
        return x*p.y-y*p.x;
    }
};
struct polygon
{
    int n;
    point p[maxp];
    void input(int _n)
    {
        n=_n;
        for (int i=0;i<n;i++)
        {
            p[i].input();
        }
    }
    void add(point q)
    {
        p[n++]=q;
    }
    struct cmp
    {
        point p;
        cmp(const point &p0){p=p0;}
        bool operator()(const point &aa,const point &bb)
        {
            point a=aa,b=bb;
            int d=dblcmp(a.sub(p).det(b.sub(p)));
            if (d==0)
            {
                return dblcmp(a.distance(p)-b.distance(p))<0;
            }
            return d>0;
        }
    };
    void getconvex(polygon &convex)
    {
        int i,j,k;
        sort(p,p+n);
        convex.n=n;
        for (i=0;i<min(n,2);i++)
        {
            convex.p[i]=p[i];
        }
        if (n<=2)return;
        int &top=convex.n;
        top=1;
        for (i=2;i<n;i++)
        {
            while (top&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
        int temp=top;
        convex.p[++top]=p[n-2];
        for (i=n-3;i>=0;i--)
        {
            while (top!=temp&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
                top--;
            convex.p[++top]=p[i];
        }
    }
    double getcircumference()
    {
        double sum=0;
        int i;
        for (i=0;i<n;i++)
        {
            sum+=p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }
};
int n;
int L;
int main()
{
    while (~scanf("%d%d",&n,&L))
    {
        polygon a,b;
        a.input(n);
        a.getconvex(b);
        double ans=b.getcircumference()+2.0*pi*L;
        printf("%0.0f\n",ans);
    }
    return 0;
}