Poj 1113 Wall

地址：http://poj.org/problem?id=1113

题目：

Wall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36064		Accepted: 12315

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

Hint

结果四舍五入就可以了

思路：凸包模板题，求出凸包边长然后加上一个圆周长就可以啦

/* 二维几何  */
/* 需要包含的头文件 */
#include <cstdio>
#include <cstring>
#include <cmath >
#include <iostream>
#include <algorithm>

using namespace std;
/** 常用的常量定义 **/
const double INF  = 1e200;
const double eps  = 1e-10;
const double PI  = acos(-1.0);
const int Max = 1e6;

/** 基本几何结构 **/
struct Point
{
    double x,y;
    Point(double a=0, double b=0){x=a,y=b;}
    bool operator<(const Point &ta)const
    {
        if(x==ta.x)     return y<ta.y;
        return x<ta.x;
    }
    friend Point operator+(const Point &ta,const Point &tb)
    {
        return Point(ta.x+tb.x,ta.y+tb.y);
    }
    friend Point operator-(const Point &ta,const Point &tb)
    {
        return Point(ta.x-tb.x,ta.y-tb.y);
    }
};
struct Vec2D        ///二维向量，*重载为点乘，/重载为叉乘
{
    double x,y;
    Vec2D(double ta,double tb){x=ta,y=tb;}
    Vec2D(Point &ta){x=ta.x,y=ta.y;}
    friend double operator*(const Vec2D &ta,const Vec2D &tb)
    {
        return ta.x*tb.x+ta.y*tb.y;
    }
    friend double operator/(const Vec2D &ta,const Vec2D &tb)
    {
        return ta.x*tb.y-ta.y*tb.x;
    }
    friend Vec2D operator+(const Vec2D &ta,const Vec2D &tb)
    {
        return Vec2D(ta.x+tb.x,ta.y+tb.y);
    }
    friend Vec2D operator-(const Vec2D &ta,const Vec2D &tb)
    {
        return Vec2D(ta.x-tb.x,ta.y-tb.y);
    }
    Vec2D operator=(const Vec2D &ta)
    {
        x=ta.x,y=ta.y;
        return *this;
    }
};
struct LineSeg      ///线段，重载了/作为叉乘运算符，*作为点乘运算符
{
    Point s,e;
    LineSeg(){s=Point(0,0),e=Point(0,0);}
    LineSeg(Point a, Point b){s=a,e=b;}
    double lenth(void)
    {
        return sqrt((s.x-e.x)*(s.x-e.x)+(s.y-e.y)*(s.y-e.y));
    }
    friend double operator*(const LineSeg &ta,const LineSeg &tb)
    {
        return (ta.e.x-ta.s.x)*(tb.e.x-tb.s.x)+(ta.e.y-ta.s.y)*(tb.e.y-tb.s.y);
    }
    friend double operator/(const LineSeg &ta,const LineSeg &tb)
    {
        return (ta.e.x-ta.s.x)*(tb.e.y-tb.s.y)-(ta.e.y-ta.s.y)*(tb.e.x-tb.s.x);
    }
    LineSeg operator=(const LineSeg &ta)
    {
        s=ta.s,e=ta.e;
        return *this;
    }
};
struct Line         /// 直线的解析方程 a*x+b*y+c=0  为统一表示，约定 a >= 0
{
    double a,b,c;
    Line(double d1=1, double d2=-1, double d3=0){ a=d1,b=d2,c=d3;}
};

double getdis(const Point &ta,const Point &tb);
bool cmp(const Point &ta,const Point &tb);
void graham(Point ps[],Point tb[],int n,int &num);
void ConvexClosure(Point ps[],Point tb[],int n,int &num);

Point ps[1100],tb[1100];
int n,l,num;
double ans;

int main(void)
{
    cin>>n>>l;
    for(int i=0,x,y;i<n;i++)
        scanf("%d%d",&x,&y),ps[i]=Point(x,y);
    graham(ps,tb,n,num);
    //for(int i=0;i<num;i++)
    //    printf("%f %f\n",tb[i].x,tb[i].y);
    tb[num++]=tb[0];
    LineSeg lx;
    ans+=2*PI*l;
    for(int i=1;i<num;i++)
    {
        lx.s=tb[i-1],lx.e=tb[i];
        ans+=lx.lenth();
    }
    printf("%d\n",(int)(0.5+ans));
    return 0;
}

double getdis(const Point &ta,const Point &tb)
{

    return sqrt((ta.x-tb.x)*(ta.x-tb.x)+(ta.y-tb.y)*(ta.y-tb.y));
}
/** ************凸包算法****************
        寻找凸包的graham 扫描法
        PS(PointSet)为输入的点集；
        tb为输出的凸包上的点集，按照逆时针方向排列;
        n为PointSet中的点的数目
        num为输出的凸包上的点的个数
****************************************** **/
bool cmp(const Point &ta,const Point &tb)
{
    double tmp=LineSeg(ps[0],ta)/LineSeg(ps[0],tb);
    if(fabs(tmp)<eps)
        return getdis(ps[0],ta)<getdis(ps[0],tb);
    else if(tmp>0)
        return 1;
    return 0;
}
void graham(Point ps[],Point tb[],int n,int &num)
{
    int cur=0,top=2;
    for(int i=1;i<n;i++)
        if(ps[cur].y>ps[i].y || (ps[cur].y==ps[i].y &&ps[cur].x>ps[i].x))
            cur=i;
    swap(ps[cur],ps[0]);
    sort(ps+1,ps+n,cmp);
    tb[0]=ps[0],tb[1]=ps[1],tb[2]=ps[2];
    for(int i=3;i<n;i++)
    {
        while(LineSeg(tb[top-1],tb[top])/LineSeg(tb[top-1],ps[i])<0)
            top--;
        tb[++top]=ps[i];
    }
    num=top+1;
}
/** 卷包裹法求点集凸壳，参数说明同graham算法 **/
void ConvexClosure(Point ps[],Point tb[],int n,int &num)
{
    LineSeg lx,ly;
    int cur,ch;
    bool vis[Max];
    num=-1,cur=0;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<n;i++)
        if(ps[cur].y>ps[i].y || (ps[cur].y==ps[i].y &&ps[cur].x>ps[i].x))
            cur=i;
    tb[++num]=ps[cur];
    lx.s=Point(ps[cur].x-1,ps[cur].y),lx.e=ps[cur];
    /// 选取与最后一条确定边夹角最小的点，即余弦值最大者
    while(1)
    {
        double mxcross=-2,midis,tmxcross;
        ly.s=lx.e;
        for(int i=0;i<n;i++)if(!vis[i])
        {
            ly.e=ps[i];
            tmxcross=(lx*ly)/lx.lenth()/ly.lenth();
            if((fabs(tmxcross-mxcross)<eps && getdis(ly.s,ly.e)<midis) || tmxcross>mxcross)
                mxcross=tmxcross,midis=getdis(ly.s,ly.e),ch=i;
        }
        if(ch==cur)break;
        tb[++num]=ps[ch],vis[ch]=1;
        lx.s=tb[num-1],lx.e=tb[num],ly.s=tb[num];
    }
    num++;
}