poj_1113 Wall（凸包）

Wall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35753		Accepted: 12201

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

题目给出的图给了提示，即求凸包的长度再加上一个以输入值L为半径的圆的周长。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define FOP2 freopen("data1.txt","w",stdout)
#define inf 0x3f3f3f3f
#define maxn 50010;
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

struct Point
{
int x, y;
Point(int x=0, int y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator - (const Point& A, const Point& B)
{
return Vector(A.x-B.x, A.y-B.y);
}

int Cross(const Vector& A, const Vector& B)
{
return A.x*B.y - A.y*B.x;
}

int Dot(const Vector& A, const Vector& B)
{
return A.x*B.x + A.y*B.y;
}

double Dist(const Point& A, const Point& B)
{
return sqrt((double)((A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y)));
}

bool operator < (const Point& p1, const Point& p2)
{
return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}

bool operator == (const Point& p1, const Point& p2)
{
return p1.x == p2.x && p1.y == p2.y;
}

// 点集凸包
// 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
// 注意：输入点集会被修改
vector<Point> ConvexHull(vector<Point>& p)
{
// 预处理，删除重复点
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());

int n = p.size();
int m = 0;
vector<Point> ch(n+1);
for(int i = 0; i < n; i++)
{
while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
ch[m++] = p[i];
}
int k = m;
for(int i = n-2; i >= 0; i--)
{
while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
ch[m++] = p[i];
}
if(n > 1) m--;
ch.resize(m);
return ch;
}

Point read_point()
{
int x, y;
scanf("%d%d", &x, &y);
return Point(x, y);
}

int n, l;
vector<Point> v;

int main()
{
while(~scanf("%d%d", &n, &l))
{
double ans = 0;
for(int i = 0; i < n; i++) v.push_back(read_point());
v = ConvexHull(v);
int m = v.size();
v.push_back(v[0]);
for(int i = 0; i < m; i++) ans += Dist(v[i], v[i+1]);
ans += (double)l*PI*2;
printf("%.f\n", ans);
}
return 0;
}