[poj1113][Wall]

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

Solution

凸包模板题，这里用水平序+上下凸壳求图包

orz clover_hxy

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAXN 3010
#define Eps 1e-18
#define Pi 3.1415926535

using namespace std;

struct Vctor{
  double x, y;

  Vctor() {}

  Vctor(double _x, double _y) : x(_x), y(_y) {}

  bool operator == (const Vctor b)const {return x == b.x && y == b.y;}

  bool operator < (const Vctor b)const {return x < b.x || (x == b.x && y < b.y);}
} d[MAXN], _pb[MAXN], e[MAXN];

Vctor operator + (Vctor a, Vctor b) {return Vctor(a.x + b.x, a.y + b.y);}

Vctor operator - (Vctor a, Vctor b) {return Vctor(a.x - b.x, a.y - b.y);}

Vctor operator * (Vctor a, double b) {return Vctor(a.x * b, a.y * b);}

Vctor operator / (Vctor a, double b) {return Vctor(a.x / b, a.y / b);}

double Dot(Vctor a, Vctor b) {return a.x * b.x + a.y * b.y;}

double Cro(Vctor a, Vctor b) {return a.x * b.y - a.y * b.x;}

int Cmp(double x){
  if(fabs(x) < Eps)return 0;
  return x < 0 ? -1 : 1;
}

double Dis(Vctor a) {return sqrt(Dot(a, a));}

double ans;
int n, L, top;

void Samsara(){
  sort(d, d + n);
  int k;
  for(int i = 0; i < n; i++){
    while(top > 1 && Cmp(Cro(_pb[top - 1] - _pb[top - 2], d[i] - _pb[top - 2])) <= 0)top--;
    _pb[top++] = d[i];
  }
  k = top;
  for(int i = n - 2; i >= 0; i--){
    while(top > k && Cmp(Cro(_pb[top - 1] - _pb[top - 2], d[i] - _pb[top - 2])) <= 0)top--;
    _pb[top++] = d[i];
  }
  if(n > 1)top--;
}

int main(){
  scanf("%d%d", &n, &L);
  for(int i = 0; i < n; i++)
    scanf("%lf%lf", &d[i].x, &d[i].y);
  Samsara();
  for(int i = 1; i <= top; i++)
    e[i] = _pb[i] - _pb[i - 1];
  e[top + 1] = _pb[0] - _pb[top];
  for(int i = 1; i <= top + 1; i++)
    ans += Dis(e[i]);
  ans += Pi * L * 2;
  printf("%.0lf\n", ans);
  return 0;
}