Naive and Silly Muggles hdu4720

Naive and Silly Muggles

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 452 Accepted Submission(s): 307

Problem 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

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <math.h>
 5 #include <algorithm>
 6 using namespace std;
 7 double p[4][2],x,y;
 8 void fun()
 9 {
10     double a=2*(p[0][1]-p[1][1]);
11     double a1=2*(p[1][1]-p[2][1]);
12     double b=2*(p[0][0]-p[1][0]);
13     double b1=2*(p[1][0]-p[2][0]);
14     double c=p[0][0]*p[0][0]+p[0][1]*p[0][1]-p[1][0]*p[1][0]-p[1][1]*p[1][1];
15     double c1=p[1][0]*p[1][0]+p[1][1]*p[1][1]-p[2][0]*p[2][0]-p[2][1]*p[2][1];
16     x=(a1*c-a*c1)/(b*a1-b1*a);
17     y=(c*b1-c1*b)/(b1*a-b*a1);
18 }
19 double dis(int k)
20 {
21     double sum;
22     sum=sqrt((x-p[k][0])*(x-p[k][0])+(y-p[k][1])*(y-p[k][1]));
23     return sum;
24 }
25 int main()
26 {
27     int t,i,j,n;
28     scanf("%d",&t);
29     for(i=1; i<=t; i++)
30     {
31         for(j=0; j<4; j++)scanf("%lf%lf",&p[j][0],&p[j][1]);
32         printf("Case #%d: ",i);
33         double a=((p[0][0]-p[1][0])*(p[0][0]-p[1][0])+(p[0][1]-p[1][1])*(p[0][1]-p[1][1]));
34         double b=((p[2][0]-p[1][0])*(p[2][0]-p[1][0])+(p[2][1]-p[1][1])*(p[2][1]-p[1][1]));
35         double c=((p[0][0]-p[2][0])*(p[0][0]-p[2][0])+(p[0][1]-p[2][1])*(p[0][1]-p[2][1]));
36         if(a*a+b*b<c*c)
37         {
38             x=(p[0][0]+p[2][0])/2;
39             y=(p[0][1]+p[2][1])/2;
40         }
41         else if(a*a+c*c<b*b)
42         {
43             x=(p[1][0]+p[2][0])/2;
44             y=(p[1][1]+p[2][1])/2;
45         }
46         else if(b*b+c*c<a*a)
47         {
48             x=(p[0][0]+p[1][0])/2;
49             y=(p[0][1]+p[1][1])/2;
50         }
51         else
52         {
53             fun();
54         }
55         double di=dis(0);
56         double di1=dis(3);
57         if(di<di1)
58         {
59             cout<<"Safe"<<endl;
60         }
61         else cout<<"Danger"<<endl;
62     }
63 }

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