Explore Track of Point

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5476

Description

In Geometry, the problem of track is very interesting. Because in some cases, the track of point may be beautiful curve. For example, in polar Coordinate system, ρ=cos3θ is like rose, ρ=1−sinθ is a Cardioid, and so on. Today, there is a simple problem about it which you need to solve.

Give you a triangle ΔABC and AB = AC. M is the midpoint of BC. Point P is in ΔABC and makes min{∠MPB+∠APC,∠MPC+∠APB} maximum. The track of P is Γ. Would you mind calculating the length of Γ?

Given the coordinate of A, B, C, please output the length of Γ.

Input

There are T (1≤T≤104) test cases. For each case, one line includes six integers the coordinate of A, B, C in order. It is guaranteed that AB = AC and three points are not collinear. All coordinates do not exceed 104 by absolute value.

Output

For each case, first please output "Case #k: ", k is the number of test case. See sample output for more detail. Then, please output the length of Γ HDU 5476 Explore Track of Point 数学平几 with exactly 4 digits after the decimal point.

Sample Input

1
0 1 -1 0 1 0

Sample Output

Case #1: 3.2214

HINT

题意

给你一个等腰三角形，底边中点叫做M，找一个点P的轨迹

使得 {∠MPB+∠APC,∠MPC+∠APB} 的最小值最大，问这个轨迹的长度是多少，这个轨迹必须在三角形内

题解：

下面这个圆弧再加上垂线的长度就好了

证明是转载的

HDU 5476 Explore Track of Point 数学平几

代码：

//qscqesze

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <bitset>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100006

#define mod 1000000007

#define eps 1e-9

#define e exp(1.0)

#define PI acos(-1)

const double EP  = 1E- ;

int Num;

//const int inf=0x7fffffff;

const ll inf=;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//*************************************************************************************

int main()

{

    int t=read();

    for(int cas=;cas<=t;cas++)

    {

        double x1,y1,x2,y2,x3,y3;

        scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);

        double a = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));

        double b = sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));

        double c = sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));

        double k = acos((a*a+b*b-c*c)/(2.0*a*b));

        double ans = cos(0.5*k)*a;

        //cout<<k<<endl;

        double h = cos(0.5*k)*a;

        double r = (a*c)/(h*2.0);

        ans += r*((double)PI-k);

        /*

        if(k<=PI/2.0)

            ans += k*a;

        else

        {

            double aa = a;

            double bb = cos(0.5*k)*a;

            double cc = c/2.0;

            double h = 2.0*bb*cc/aa;

            double kk = acos(h/aa);

            double kkk = k - 4.0*kk;

            kkk = max(0.0,kkk);

            ans += kkk*a;

        }

        */

        printf("Case #%d: %.4lf\n",cas,ans);

    }

}

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