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

Output

Sample Input

9 100200 400300 400300 300400 300400 400500 400500 200350 200200 200

Sample Output

1628

Hint

Source

求出围住这个城堡的围墙的最小值，围墙距离城堡最低不能小于l

所以结果= 城堡的凸包边长 + l的圆周长。

在这个题中求凸包的时候，进行极角排序->压入节点->分类如果mul(q2,q1,p[i]) = 1 直接压入->= 0 抛出栈首的节点，压入新节点-> = -1只抛出栈首的节点i--。

#include <cstdio>#include <cstring>#include <algorithm>#include <stack>#include <cmath>using namespace std ;#define eps 1e-8#define PI 3.141592653589793238462643383279502884197169399375105820974944struct node{    double x , y ;}p[11000] , q , q1 , q2 ;stack <node> sta ;node sub(node a,node b){    a.x -= b.x ;    a.y -= b.y ;    return a ;}int mul(node q,node a,node b){    a = sub(a,q) ;    b = sub(b,q) ;    double x = a.x*b.y - a.y*b.x ;    if( fabs(x) < eps )        return 0 ;    else if( x > 0 )        return 1 ;    return -1 ;}int dis(node q,node a,node b){    a = sub(a,q) ;    b = sub(b,q) ;    double k1 = a.x*a.x + a.y*a.y , k2 = b.x*b.x + b.y*b.y ;    if( fabs(k1-k2) < eps )        return 0 ;    else if( k1-k2 > eps )        return 1 ;    return -1 ;}int cmp(node a,node b){    int k1 = mul(q,a,b) , k2 = dis(q,a,b) ;    return k1 == 1 || (k1 == 0 && k2 == -1) ;}int main(){    int n , i , j ;    double r , ans ;    while( scanf("%d %lf", &n, &r) != EOF )    {        q.x = q.y = 11000.0 ;        for(i = 0 ; i < n ; i++)        {            scanf("%lf %lf", &p[i].x, &p[i].y) ;            if( p[i].y < q.y || ( fabs(p[i].y-q.y ) < eps && p[i].x < q.x ) )                q = p[i] ;        }        sort(p,p+n,cmp) ;        while( !sta.empty() ) sta.pop() ;        sta.push(p[0]) ;        sta.push(p[1]) ;        for(i = 2 ; i < n ; i++)        {            //printf("%lf %lf*\n", p[i].x, p[i].y) ;            q1 = sta.top() ;            sta.pop() ;            q2 = sta.top() ;            sta.pop() ;            int k = mul(q2,q1,p[i]) ;            if( k == 1 )            {                sta.push(q2) ;                sta.push(q1) ;                sta.push(p[i]) ;            }            else if( k == 0 )            {                sta.push(q2) ;                sta.push(p[i]) ;            }            else            {                sta.push(q2) ;                i-- ;            }        }        n = 1 ;        while( !sta.empty() )        {            p[n++] = sta.top() ;            //printf("%lf %lf\n", p[n-1].x, p[n-1].y);            sta.pop() ;        }        ans = 2.0*PI*r ;        for(i = 0 ; i < n-1 ; i++)        {            ans += sqrt( (p[i].x-p[i+1].x)*(p[i].x-p[i+1].x)*1.0 + (p[i].y-p[i+1].y)*(p[i].y-p[i+1].y)*1.0 ) * 1.0 ;        }        printf("%.0f\n", ans) ;    }    return 0;}

poj1113--Wall(凸包)