ACM学习历程—HDU4720 Naive and Silly Muggles(计算几何）

Description

Three wizards are doing a experiment. To avoid from bothering, a special magic is set around them. The magic forms a circle, which covers those three wizards, in other words, all of them are inside or on the border of the circle. And due to save the magic power, circle's area should as smaller as it could be.
Naive and silly "muggles"(who have no talents in magic) should absolutely not get into the circle, nor even on its border, or they will be in danger.
Given the position of a muggle, is he safe, or in serious danger?

 

Input

The first line has a number T (T <= 10) , indicating the number of test cases.
For each test case there are four lines. Three lines come each with two integers x _i and y _i (|x _i, y _i| <= 10), indicating the three wizards' positions. Then a single line with two numbers q _x and q _y (|q _x, q _y| <= 10), indicating the muggle's position.

 

Output

For test case X, output "Case #X: " first, then output "Danger" or "Safe".

 

Sample Input

3

0 0

2 0

1 2

1 -0.5

 

0 0

2 0

1 2

1 -0.6

 

0 0

3 0

1 1

1 -1.5

 

Sample Output

Case #1: Danger

Case #2: Safe

Case #3: Safe

 

题目大意就是先求一个能包含三个点的最小圆，然后判断第四个圆是否在圆内。

这三点中取出两点，如果以这两个点构成的线段为直径，能包含第三个点，自然便是最小圆。于是先考虑最远的两个点即可。

其次，如果上述不满足（三点一线的满足上面），自然需要逐步扩大直径来包含第三个点，自然所求的便是外接圆。

 

对于求外接圆，此处采用了暴力设圆心坐标(x, y)

所以(x-x1)^2 + (y-y1)^2 = (x-x2)^2 + (y-y2)^2 = (x-x3)^2 + (y-y3)^2

化简得到：

 

2*((x1-x2)*(y1-y3) - (x1-x3)*(y1-y2)) * x

= (y1-y2)*(y2-y3)*(y1-y3) + (x1*x1-x2*x2)*(y1-y3) - (x1*x1-x3*x3)*(y1-y2);

 

2*((y1-y2)*(x1-x3) - (y1-y3)*(x1-x2)) * y

= (x1-x2)*(x2-x3)*(x1-x3) + (y1*y1-y2*y2)*(x1-x3) - (y1*y1-y3*y3)*(x1-x2);

 

于是圆心求出来问题便简单了。

 

代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#define LL long long



using namespace std;



double x1,x2,x3,y1,y2,y3, x0, y0;

double rx, ry, r2;

int  n,i;



void Cal()

{

    double A, B;

    A = 2*((x1-x2)*(y1-y3) - (x1-x3)*(y1-y2));

    B = (y1-y2)*(y2-y3)*(y1-y3) + (x1*x1-x2*x2)*(y1-y3) - (x1*x1-x3*x3)*(y1-y2);

    rx = B/A;



    A = 2*((y1-y2)*(x1-x3) - (y1-y3)*(x1-x2));

    B = (x1-x2)*(x2-x3)*(x1-x3) + (y1*y1-y2*y2)*(x1-x3) - (y1*y1-y3*y3)*(x1-x2);

    ry = B/A;

    r2 = (rx-x1)*(rx-x1) + (ry-y1)*(ry-y1);

}



void Work()

{

    int cnt = 3;

    double tmp;

    r2 = ((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))/4;

    rx = (x1+x2)/2;

    ry = (y1+y2)/2;

    tmp = ((x2-x3)*(x2-x3) + (y2-y3)*(y2-y3))/4;

    if (tmp > r2)

    {

        cnt = 1;

        r2 = tmp;

        rx = (x3+x2)/2;

        ry = (y3+y2)/2;

    }

    tmp = ((x1-x3)*(x1-x3) + (y1-y3)*(y1-y3))/4;

    if (tmp > r2)

    {

        cnt = 2;

        r2 = tmp;

        rx = (x1+x3)/2;

        ry = (y1+y3)/2;

    }

    switch (cnt)

    {

        case 1:

            tmp = (rx-x1)*(rx-x1) + (ry-y1)*(ry-y1);

            break;

        case 2:

            tmp = (rx-x2)*(rx-x2) + (ry-y2)*(ry-y2);

            break;

        case 3:

            tmp = (rx-x3)*(rx-x3) + (ry-y3)*(ry-y3);

            break;

    }

    if (tmp > r2)

    {

        Cal();

    }

}



void Output()

{

    if (r2 >= (rx-x0)*(rx-x0) + (ry-y0)*(ry-y0))

        printf("Danger\n");

    else

        printf("Safe\n");

}



int main()

{

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

    int T;

    scanf("%d", &T);

    for(int times = 1; times <= T; times++)

    {

        scanf("%lf%lf", &x1, &y1);

        scanf("%lf%lf", &x2, &y2);

        scanf("%lf%lf", &x3, &y3);

        scanf("%lf%lf", &x0, &y0);

        Work();

        printf("Case #%d: ", times);

        Output();

    }

}