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
геометрия – выпуклая оболочка
Периметр стены равен длине
выпуклой оболочки плюс длина окружности радиуса L.
Реализация алгоритма
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#define PI
acos(-1.0)
using namespace std;
class Point
{
public:
int x, y;
Point(int x =
0, int y = 0)
{
this->x
= x; this->y = y;
}
double len2()
const {return
x*x + y*y;}
};
Point operator+ (Point a, Point b)
{
return
Point(a.x+b.x,a.y+b.y);
}
Point operator- (Point a, Point b)
{
return
Point(a.x-b.x,a.y-b.y);
}
vector<Point>
v, hull;
int i, n,
cur, a, b;
double p, l;
int
f(Point a, Point b)
{
Point q = a - v[0], w = b - a;
if(q.x * w.y
> w.x * q.y)
return 1;
if(q.x * w.y
== w.x * q.y)
return
a.len2() < b.len2();
return 0;
}
int
TurnLeft(Point a, Point b, Point c)
{
Point q = b - a, w = c - b;
return q.x *
w.y > w.x * q.y;
}
int main(void)
{
scanf("%d
%lf",&n,&l);
for(i = 0; i
< n; i++)
{
scanf("%d
%d",&a,&b);
v.push_back(Point(a,b));
}
for(i = 1; i
< n; i++)
{
if (v[i].x
< v[0].x) swap(v[i],v[0]);
if ((v[i].x
== v[0].x) && (v[i].y < v[0].y)) swap(v[i],v[0]);
}
sort(v.begin()+1,v.end(),f); v.push_back(v[0]); n++;
for(cur = 1,
i = 2; i < n; i++)
{
while(
(cur > 0) && !TurnLeft(v[cur-1],v[cur],v[i]))
cur--;
v[++cur] = v[i];
}
for(i = 0; i
< cur; i++)
p += sqrt(1.0*(v[i+1] - v[i]).len2());
p += 2 * PI * l;
printf("%.0lf\n",p);
return 0;
}