POJ 1113 Wall （凸包）

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 100200 400300 400300 300400 300400 400500 400500 200350 200200 200

Sample Output

1628

Hint

结果四舍五入就可以了

题目大意为国王让一个建筑师建造一个城墙，城墙距离城堡的距离不能小于L，求城墙的最小周长。这就要求在拐角处是以L为半径的1/4圆，一共有四个拐角处，所以正好所构成一个圆。所以很明显本题要先构造一个凸包，城墙的距离= 凸包的周长 + 圆的周长。关于凸包的构造在此就不在赘述，下面见代码：

#include <stdio.h>#include <cmath>#include<cstring>#include <algorithm>#define N 1005using namespace std;const double PI = acos(-1.0);struct point{int x,y;}p[N];int n,l,s[N],top;int cross(point a, point b, point c){ return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);}double dis(point a, point b){  return sqrt((b.x - a.x) * (b.x - a.x) + (b.y - a.y) * (b.y - a.y));}bool cmp(point a, point b){ int ans = cross(p[0],a,b);if(ans > 0 || (!ans && dis(p[0],a) > dis(p[0],b) ))return true;return false;}void Graham(){if(n == 1){top = 0;s[0] = 0;}else if(n >= 2){top = 1;s[0] = 0;s[1] = 1;for(int i = 2; i < n; i ++){while(top && cross(p[s[top - 1]],p[s[top]],p[i]) < 0) top --;s[++ top] = i;}}}void RC(){double ans = 0.;for(int i = 0; i < top; i ++)ans += dis(p[s[i]],p[s[i + 1]]);ans += dis(p[s[0]],p[s[top]]);ans += 2 * PI * l;printf("%d\n",(int)(ans + 0.5));}int main(){while(~scanf("%d%d",&n,&l)){for(int i = 0; i < n; i ++)scanf("%d%d",&p[i].x,&p[i].y);int u = 0;for(int i = 1; i < n; i ++){if(p[u].y > p[i].y || (p[u].y == p[i].y && p[u].x > p[i].x))u = i;}if(u){swap(p[u].y,p[0].y);swap(p[u].x,p[0].x);}sort(p + 1, p + n, cmp);Graham();RC();}return 0;}