POJ 1113 Wall （凸包）

Wall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21502		Accepted: 7048

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

結果四舍五入就可以了

Source

Northeastern Europe 2001

 

 

這道題的答案是凸包周長加上一個圓周長，即包圍凸包的一個圓角多邊形，但是沒弄明白那些圓角加起來為什么恰好是一個圓。每個圓角是以凸包對應的頂點為圓心，給定的L為半徑，與相鄰兩條邊的切點之間的一段圓弧。每個圓弧的兩條半徑夾角與對應的凸包的內角互補。假設凸包有n條邊，則所有圓弧角之和為180°*n-180°*（n-2）=360°。故，圍牆周長為=n條平行於凸包的線段+n條圓弧的長度=凸包周長+圍牆離城堡距離L為半徑的圓周長。

 

程序如下：

自己寫的凸包模版，Graham,細節處修改了下。

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=1000;
const double PI=acos(-1.0);

struct point
{
    int x,y;
};
point list[MAXN];
int stack[MAXN],top;

int cross(point p0,point p1,point p2) //計算叉積  p0p1 X p0p2 
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}    
double dis(point p1,point p2)  //計算 p1p2的 距離 
{
    return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}    
bool cmp(point p1,point p2) //極角排序函數 ， 角度相同則距離小的在前面 
{
    int tmp=cross(list[0],p1,p2);
    if(tmp>0) return true;
    else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;
    else return false;
}    
void init(int n) //輸入，並把  最左下方的點放在 list[0]  。並且進行極角排序 
{
    int i,k;
    point p0;
    scanf("%d%d",&list[0].x,&list[0].y);
    p0.x=list[0].x;
    p0.y=list[0].y;
    k=0;
    for(i=1;i<n;i++)
    {
        scanf("%d%d",&list[i].x,&list[i].y);
        if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].x)) )
        {
            p0.x=list[i].x;
            p0.y=list[i].y;
            k=i;
        }    
    }    
    list[k]=list[0];
    list[0]=p0;
    
    sort(list+1,list+n,cmp);
}     

void graham(int n)
{
    int i;
    if(n==1) {top=0;stack[0]=0;}
    if(n==2)
    {
        top=1;
        stack[0]=0;
        stack[1]=1;
    }    
    if(n>2)
    {
        for(i=0;i<=1;i++) stack[i]=i;
        top=1;
        
        for(i=2;i<n;i++)
        {
            while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) top--;
            top++;
            stack[top]=i;
        }    
    }    
}    

int main()
{
    int N,L;
    while(scanf("%d%d",&N,&L)!=EOF)
    {
        init(N);
        graham(N);
        double res=0;
        for(int i=0;i<top;i++)
          res+=dis(list[stack[i]],list[stack[i+1]]);
        res+=dis(list[stack[0]],list[stack[top]]);
        
        res+=2*PI*L;
        printf("%d\n",(int)(res+0.5));
        
    }    
    return 0;
}