Naive and Silly Muggles
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24 Accepted Submission(s): 17
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
Source
Recommend
zhuyuanchen520
首先是找出一个最小的圆来覆盖三个点。
圆心要么是三条线段的中点，或者三角形的外心，很容易找出来。‘
然后判断点是否在圆外就可以了
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-9-11 13:17:28 4 File Name :2013-9-111005.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 const double eps = 1e-8; 21 int sgn(double x) 22 { 23 if(fabs(x) < eps)return 0 ; 24 if ( x < 0)return -1; 25 else return 1; 26 } 27 struct Point 28 { 29 double x,y; 30 Point(double _x = 0, double _y = 0) 31 { 32 x = _x; 33 y = _y; 34 } 35 Point operator -(const Point &b)const 36 { 37 return Point(x-b.x,y-b.y); 38 } 39 double operator ^(const Point &b)const 40 { 41 return x*b.y - y*b.x; 42 } 43 void input() 44 { 45 scanf("%lf%lf",&x,&y); 46 } 47 }; 48 double dist(Point a,Point b) 49 { 50 return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y)); 51 } 52 53 //过三点求圆心坐标 54 Point waixin(Point a,Point b,Point c) 55 { 56 double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2; 57 double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2; 58 double d = a1*b2 - a2*b1; 59 return Point(a.x + (c1*b2 - c2*b1)/d, a.y + (a1*c2 -a2*c1)/d); 60 } 61 62 63 int main() 64 { 65 //freopen("in.txt","r",stdin); 66 //freopen("out.txt","w",stdout); 67 int T; 68 int iCase = 0; 69 Point p[4]; 70 scanf("%d",&T); 71 while(T--) 72 { 73 iCase++; 74 for(int i = 0;i < 4;i++) 75 p[i].input(); 76 Point res; 77 double tmp = 1e20; 78 for(int i = 0;i < 3;i++) 79 { 80 Point t = Point((p[i].x+p[(i+1)%3].x)/2,(p[i].y+p[(i+1)%3].y)/2); 81 double dd = max(dist(p[0],t),max(dist(p[1],t),dist(p[2],t))); 82 if(dd < tmp) 83 { 84 tmp = dd; 85 res = t; 86 } 87 } 88 if(sgn( (p[1]-p[0])^(p[2]-p[0]) ) != 0) 89 { 90 Point t = waixin(p[0],p[1],p[2]); 91 double dd = max(dist(p[0],t),max(dist(p[1],t),dist(p[2],t))); 92 if(dd < tmp) 93 { 94 tmp = dd; 95 res = t; 96 } 97 98 } 99 printf("Case #%d: ",iCase); 100 if(sgn(tmp - dist(res,p[3])) >= 0) 101 printf("Danger "); 102 else printf("Safe "); 103 } 104 return 0; 105 }