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?
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 xi and yi (|xi, yi| <= 10), indicating the three wizards' positions. Then a single line with two numbers qx and qy (|qx, qy| <= 10), indicating the muggle's position.
For each test case there are four lines. Three lines come each with two integers xi and yi (|xi, yi| <= 10), indicating the three wizards' positions. Then a single line with two numbers qx and qy (|qx, qy| <= 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 }