Sicily 1012. Stacking Cylinders 解题报告

题目地址：Sicily 1012. Stacking Cylinders

思路：

　　最低层如果有n个圆，则一共会有n层，其中最高一层有1个。用n个数组记录n层圆的坐标，一开始输入底层的坐标，排序之后再不断利用下一层的坐标算出当前一层的坐标，知道最高层就行。对于每一个圆心坐标，可以用支撑它的下面两个圆心坐标通过几何计算方法算出。注意输出的格式。

代码：

#include<iostream>
#include<iomanip>
#include<cmath>
#include<algorithm>
using namespace std;

struct Point{
    double x,y;
};

void get_a_point_from_two_below(Point &p,const Point &p1,const Point &p2);
bool cmp(const Point &p1,const Point &p2){
    return p1.x<p2.x;
}

int main(){
    int number_of_cylinders;
    while(cin>>number_of_cylinders&&number_of_cylinders!=0){
        Point *level[number_of_cylinders];
        for(int i=0;i<number_of_cylinders;i++) //create room for each level of cylinders.
            level[i]=new Point[number_of_cylinders-i];
        for(int i=0;i<number_of_cylinders;i++){ //get data of the bottom level
            cin>>level[0][i].x;
            level[0][i].y=1;
        }
        sort(level[0],level[0]+number_of_cylinders,cmp);
        for(int i=1;i<number_of_cylinders;i++){
            for(int j=0;j<number_of_cylinders-i;j++){
                get_a_point_from_two_below(level[i][j],level[i-1][j],
                        level[i-1][j+1]);
            }
        }
        cout.setf(ios::fixed);
        cout<<setprecision(4)<<level[number_of_cylinders-1][0].x<<' '
                <<level[number_of_cylinders-1][0].y<<endl;
        for(int i=0;i<number_of_cylinders;i++)
            delete level[i];
    }
    return 0;
}
void get_a_point_from_two_below(Point &p,const Point &p1,const Point &p2){
    double side=sqrt(pow(p2.x-p1.x,2)+pow(p2.y-p1.y,2));
    p.x=p1.x+2*cos(acos(side/4)+atan((p2.y-p1.y)/(p2.x-p1.x)));
    p.y=p1.y+2*sin(acos(side/4)+atan((p2.y-p1.y)/(p2.x-p1.x)));
}