ural 1075. Thread in a Space

1075. Thread in a Space

Time limit: 1.0 second
Memory limit: 64 MB

There are three points in a 3-dimensional space: A, B and C. All the coordinates of the points are integer numbers with absolute values not greater than 1000. A solid ball with a center in the pointC is firmly fixed. A radius of the ball is R, a positive integer number. Distances from the point Cto the points A and B are strictly greater than R.

It is necessary to stretch a thread of minimal length between points A and B. Surely, the thread should be outside of the ball.

You are to find out a length of the thread.

Input

The first three lines contain coordinates of the points A, B and C respectively. The fourth one contains a radius R of the ball.

Output

should contain a minimal length of the thread to within 2 symbols after a decimal point. You should output answer with two or more digits.

Sample

input	output
0 0 12 12 0 0 10 0 10 10	19.71

Problem Author: Alexander Mironenko
Problem Source: Ural State Univerisity Personal Contest Online February'2001 Students Session

Tags: geometry (hide tags for unsolved problems)

Difficulty: 1029

题意：给出空间中三个整点。在一点的中央，固定了一个小球。小球的半径为R。其余两点在球外。用最小长度的绳子连接剩下两点。求绳子的长度。

分析：因为只有三个点，所以可以放在平面上考虑。

如果连线不经过圆，显然可以直接求距离。

如果经过，那么有简单的平面几何知识即可。

而且处理时不需理会什么空间，直接用角度算即可。

 /**

 Create By yzx - stupidboy

 */

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <cmath>

 #include <deque>

 #include <vector>

 #include <queue>

 #include <iostream>

 #include <algorithm>

 #include <map>

 #include <set>

 #include <ctime>

 #include <iomanip>

 using namespace std;

 typedef long long LL;

 typedef double DB;

 #define MIT (2147483647)

 #define INF (1000000001)

 #define MLL (1000000000000000001LL)

 #define sz(x) ((int) (x).size())

 #define clr(x, y) memset(x, y, sizeof(x))

 #define puf push_front

 #define pub push_back

 #define pof pop_front

 #define pob pop_back

 #define ft first

 #define sd second

 #define mk make_pair

 inline int Getint()

 {

     int Ret = ;

     char Ch = ' ';

     bool Flag = ;

     while(!(Ch >= '' && Ch <= ''))

     {

         if(Ch == '-') Flag ^= ;

         Ch = getchar();

     }

     while(Ch >= '' && Ch <= '')

     {

         Ret = Ret *  + Ch - '';

         Ch = getchar();

     }

     return Flag ? -Ret : Ret;

 }

 const DB EPS = 1e-;

 struct Point

 {

     DB x, y, z;

     inline void Read()

     {

         cin >> x >> y >> z;

     }

 } a, b, c;

 DB r;

 inline void Input()

 {

     a.Read();

     b.Read();

     c.Read();

     cin >> r;

 }

 inline DB Sqr(DB x)

 {

     return x * x;

 }

 inline DB Dist(Point a, Point b)

 {

     return sqrt(Sqr(a.x - b.x) + Sqr(a.y - b.y) + Sqr(a.z - b.z));

 }

 inline void Solve()

 {

     DB ans = 0.0;

     DB  ab = Dist(a, b),

         ac = Dist(a, c),

         bc = Dist(b, c);

     DB angle =

         acos((Sqr(ac) + Sqr(bc) - Sqr(ab)) / (2.0 * ac * bc)) -

         acos(r / ac) - acos(r/ bc);

     if(angle <= EPS)

     {

         printf("%.2lf\n", ab);

         return;

     }

     ans += sqrt(Sqr(ac) - Sqr(r));

     ans += sqrt(Sqr(bc) - Sqr(r));

     ans += angle * r;

     printf("%.2lf\n", ans);

 }

 int main()

 {

     freopen("g.in", "r", stdin);

     Input();

     Solve();

     return ;

 }

巴特西

ural 1075. Thread in a Space

1075. Thread in a Space

Input

Output

Sample

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