A - Wall(凸包Andrew模板)

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.

A - Wall(凸包Andrew模板)_第1张图片


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.

Sample

Inputcopy Outputcopy
9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200
1628

题意:

 给一个多边形,求凸包,因为它让边离顶点距离最小是L,在每个顶点画一个以l为半径的圆,会发现墙的周长恰好为:凸包周长+以l为半径圆的周长。

代码:

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
#define per(i,a,b) for(int i=a;i<=b;i++)
#define ber(i,a,b) for(int i=a;i>=b;i--)
const int N = 1e5 + 1000;
const double eps = 1e-8;
double PI = (double)acos(-1*1.0);
struct Point
{
    double x;
    double y;
}p[N],s[N];
int n, top = 0,l;
bool cmp(struct Point& a, struct Point& b)
{
    if (a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}
double cross(struct Point& a, struct Point& b, struct Point& c)
{
    return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
double dis(struct Point& a, struct Point& b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double Andrew()
{
    sort(p + 1, p + 1 + n, cmp);
    per(i, 1, n)
    {
        while (top > 1 && cross(s[top - 1], s[top], p[i]) <= 0) top--;
        s[++top] = p[i];
    }
    int t = top;
    ber(i, n - 1, 1)
    {
        while (top > t && cross(s[top - 1], s[top], p[i]) <= 0)top--;
        s[++top] = p[i];
    }
    double ans = 0;
    per(i, 1, top - 1)
        ans += dis(s[i], s[i + 1]);
    return ans;
}
int main()
{
    cin >> n >> l;
    per(i, 1, n)
   cin >> p[i].x >> p[i].y;
    double ans = Andrew();
    ans += 2 * PI * l;
    if (ans - LL(ans) >= 0.5)
        ans++;
    cout << (LL)ans << endl;

    
    return 0;
}

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