B. Rolling The Polygon

  Bahiyyah has a convex polygon with n vertices P0, P1, · · · , Pn−1 in the counterclockwiseorder. Two vertices with consecutive(连续的)  indexes are adjacent(邻近的), and besides, P0 and Pn−1 are adjacent. She also assigns(指出) a point Q inside the polygon which may appear on the border.

  Now, Bahiyyah decides to roll the polygon along a straight line and calculate the length of the trajectory(弹道) (or track) of point Q.

  To help clarify, we suppose Pn = P0, Pn+1 = P1 and assume the edge between P0 and P1 is lying on the line at first. At that point when the edge between Pi−1 and Pi lies on the line, Bahiyyah rolls the polygon forward rotating(旋转) the polygon along the vertex Pi until the next edge (which is between Pi and Pi+1) meets the line. She will stop the rolling when the edge between Pn and Pn+1 (which is same as the edge between P0 and P1) meets the line again.

 

Input

The input contains several test cases, and the first line is a positive integer T indicating the number of test cases which is up to 50.

For each test case, the first line contains an integer n (3 ≤ n ≤ 50) indicating the number of vertices of the given convex polygon. Following n lines describe vertices of the polygon in the counterclockwise order. The i-th line of them contains two integers xi−1 and yi−1, which are the coordinates of point Pi−1. The last line contains two integers xQ and yQ, which are the coordinates of point Q.

We guarantee that all coordinates are in the range of −103 to 103, and point Q is located inside the polygon or lies on its border.

Output

For each test case, output a line containing Case #x: y, where x is the test case number starting from 1, and y is the length of the trajectory of the point Q rounded to 3 places. We guarantee that 4-th place after the decimal point in the precise answer would not be 4 or 5.  

题目理解

  题目描述很短，大意是现在有一个n边的凸多边形，给你这个多边形所有顶点的坐标。然后再给你多边形内一个点的坐标，接着滚动这个多边形一周，需要我们计算出目标点移动轨迹的长度。为了更好理解，题目第三段详细说明了如何滚动：假设P0为初始点，那么在P0和P1之间有一条边1，在P1和P2之间有一条边2，连接P0和P2也有一条边3，每次都沿着一条边滚动，且第一次都是沿着边1滚。如何停止呢？就是当滚动的边再一次是边1时就停止。可以认为最后是Pn和Pn+1之间的边，即Pn=P0,Pn+1=P1。

解题思路

  观察多边形和目标点滚动的方式和轨迹，可以知道每次滚动所形成的轨迹是一段弧长，那么为了求出此弧长则需要知道此弧长所在圆的半径和旋转的角度。对于半径，首先确定圆心即使滚动角的顶点（滚动角就是Pi、Pi+1和Pi+2所形成的夹角，那么滚动角顶点则是P1），而目标点则是圆边上的一点，那么半径= 滚动角顶点(v[i+1])到目标点(tv)的距离 = sqrt(pow(tv.x-ver[i+1].x,2)+pow(tv.y-ver[i+1].y,2))；对于旋转角度就是180°减去滚动角的角度，三点坐标有了,Pi和Pi+1之间边a，那么就用余弦定理（复习下：cosx = (a^2 + b^2 - c^2)/(2ab)）。旋转角度= 180°-滚动角° = π-acos((a*a+b*b-c*c)/(2*a*b)) ，（注：Pi和Pi+1之间边a，Pi+1和Pi+2之间边b，Pi和Pi+2之间边c）；最后求弧长 = 半径 * 旋转角度即可。

  弧长计算的循环如果从P0开始，Pn-1结束，就设Pn=P0，Pn+1=P1，因为根据题意的停止条件（当边再次滚到初始的边1时，滚完即止，最后是Pn-1要和Pn和Pn+1构成一个角，那就要设Pn=P0，Pn+1=P1）。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#define inf 0x3f3f3f3f
const int maxn=1e5+10;
using namespace std;
typedef long long ll;
int x[100],y[100],xx,yy;
double dis[100],u[100];
#define PI acos(-1.0)


double calu(int a,int b)//计算半径
{
    return sqrt((a-xx)*(a-xx)+(b-yy)*(b-yy));
}
double caluu(int a,int b,int c,int d)
{
    return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}

int main()
{
    int T;
    cin>>T;
    for(int j=1;j<=T;j++)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>x[i]>>y[i];
        }
        cin>>xx>>yy;
        for(int i=1;i<=n;i++)
        {
            dis[i]=calu(x[i],y[i]);
        }
        double t1,t2,t3;
        for(int i=1;i<=n;i++)
        {
            if(i==1)
            {
                t1=caluu(x[1],y[1],x[n],y[n]);
                t2=caluu(x[1],y[1],x[2],y[2]);
                t3=caluu(x[n],y[n],x[2],y[2]);
            }
            else if(i==n)
            {
                t1=caluu(x[n],y[n],x[1],y[1]);
                t2=caluu(x[n],y[n],x[n-1],y[n-1]);
                t3=caluu(x[1],y[1],x[n-1],y[n-1]);
            }
            else
            {
                t1=caluu(x[i],y[i],x[i+1],y[i+1]);
                t2=caluu(x[i],y[i],x[i-1],y[i-1]);
                t3=caluu(x[i-1],y[i-1],x[i+1],y[i+1]);
            }
            u[i]=acos((t1*t1+t2*t2-t3*t3)/(2*t1*t2)); //计算夹角
        }
        double sum=0;
        for(int i=1;i<=n;i++)
        {
            sum+=dis[i]*(PI-u[i]);//弧长 = 半径 * 旋转角度（180度-夹角）
        }
        printf("Case #%d: %.3lf\n",j,sum);
    }
    return 0;
}