HDUOJ-------Naive and Silly Muggles
Naive and Silly Muggles
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 61 Accepted Submission(s): 39
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 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
几何题：
考虑的事情有：
(1)三点是否在一条直线上...求出前后坐标，得出圆心，和半径r;
(2)区分锐角和钝角三角形....锐角三角形（最小的圆为其外接圆），钝角三角形以最长边为直径做圆为其最小圆面积...
于是 有一点必须要注意，那就是求 外接圆的中心坐标（x,y）
代码wei:
1 通俗算法
2 定义：设平面上的三点A(x1,y1),B(x2,y2),C(x3,y3),定义
3 S(A,B,C) = (x1-x3)*(y2-y3) - (y1-y3)*(x2-x3)
4
5 已知三角形的三个顶点为A(x1,y1),B(x2,y2),C(x3,y3)，则该三角形的外心为：
6 S((x1*x1+y1*y1, y1), (x2*x2+y2*y2, y2), (x3*x3+y3*y3, y3))
7 x0 = -----------------------------------------------------------
8 2*S(A,B,C)
9
10 S((x1,x1*x1+y1*y1), (x2, x2*x2+y2*y2), (x3, x3*x3+y3*y3))
11 y0 = -----------------------------------------------------------
12 2*S(A,B,C) 代码形式：
1 //求外接圆的圆心
2 double S(double x1,double y1,double x2,double y2,double x3,double y3){
3 return ((x1-x3)*(y2-y3) - (y1-y3)*(x2-x3) );
4 }
5
6 double getx(double x1,double y1,double x2,double y2,double x3,double y3){
7 return (S(x1*x1+y1*y1,y1, x2*x2+y2*y2, y2,x3*x3+y3*y3,y3)/(2*S(x1,y1,x2,y2,x3,y3)) );
8 }
9
10 double gety(double x1,double y1,double x2,double y2,double x3,double y3){
11 return (S(x1, x1*x1+y1*y1, x2, x2*x2+y2*y2, x3, x3*x3+y3*y3) / (2*S(x1,y1,x2,y2,x3,y3)));
12 }Sample Output
Case #1: Danger
Case #2: Safe
Case #3: Safe
此题代码为：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 using namespace std;
6 bool isline(double *a,double *b,double *c)
7 {
8 if(fabs((b[1]-a[1])*(c[0]-a[0])-(c[1]-a[1])*(b[0]-a[0]))<1e-8)
9 return 1;
10 else
11 return 0;
12 }
13 //求外接圆的圆心
14 double S(double x1,double y1,double x2,double y2,double x3,double y3){
15 return ((x1-x3)*(y2-y3) - (y1-y3)*(x2-x3) );
16 }
17
18 double getx(double x1,double y1,double x2,double y2,double x3,double y3){
19 return (S(x1*x1+y1*y1,y1, x2*x2+y2*y2, y2,x3*x3+y3*y3,y3)/(2*S(x1,y1,x2,y2,x3,y3)) );
20 }
21
22 double gety(double x1,double y1,double x2,double y2,double x3,double y3){
23 return (S(x1, x1*x1+y1*y1, x2, x2*x2+y2*y2, x3, x3*x3+y3*y3) / (2*S(x1,y1,x2,y2,x3,y3)));
24 }
25 //求两条边的夹角
26 bool iftrue(double *a,double *b,double *c )
27 {
28 return (a[0]-b[0])*(c[0]-b[0])+(a[1]-b[1])*(c[1]-b[1])>0?0:1; //不是锐角时yes
29 }
30 //求两点间的距离
31 double distan(double *a,double *b)
32 {
33 return sqrt((a[0]-b[0])*(a[0]-b[0])+(a[1]-b[1])*(a[1]-b[1]))/2.0;
34 }
35
36 int main()
37 {
38 int t,count,i;
39 double po[4][2],r,save[2][2],x,y;
40 scanf("%d",&t);
41 for(count=1;count<=t;count++)
42 {
43 for(i=0;i<4;i++)
44 {
45 scanf("%lf%lf",&po[i][0],&po[i][1]);
46 if(i==0||save[1][0]*save[1][0]+save[1][1]*save[1][1]<po[i][0]*po[i][0]+po[i][1]*po[i][1])
47 save[1][1]=po[i][1],save[1][0]=po[i][0];
48 if(i==0||save[0][0]*save[0][0]+save[0][1]*save[0][1]>po[i][0]*po[i][0]+po[i][1]*po[i][1])
49 save[0][1]=po[i][1],save[0][0]=po[i][0];
50 }
51 if(isline(po[0],po[1],po[2]))
52 {
53 r=sqrt((save[1][0]-save[0][0])*(save[1][0]-save[0][0])+(save[1][1]-save[0][1])*(save[1][1]-save[0][1]))/2.0;
54 x=(save[0][0]+save[1][0])/2.0;
55 y=(save[0][1]+save[1][1])/2.0;
56 }
57 else
58 {
59 bool judge[3];
60 judge[0]=iftrue(po[0],po[1],po[2]);
61 judge[1]=iftrue(po[1],po[0],po[2]);
62 judge[2]=iftrue(po[1],po[2],po[0]);
63 if(judge[0]||judge[1]||judge[2])
64 {
65 if(judge[0])
66 {
67 x=(po[0][0]+po[2][0])/2.0;
68 y=(po[0][1]+po[2][1])/2.0;
69 r=distan(po[0],po[2]);
70 }
71 else if(judge[1])
72 {
73 x=(po[1][0]+po[2][0])/2.0;
74 y=(po[1][1]+po[2][1])/2.0;
75 r=distan(po[1],po[2]);
76 }
77 else if(judge[2])
78 {
79 x=(po[1][0]+po[0][0])/2.0;
80 y=(po[1][1]+po[0][1])/2.0;
81 r=distan(po[0],po[1]);
82 }
83 }
84 else
85 {
86 //当为锐角时，求其外接圆，否者不求
87 x=getx(po[0][0],po[0][1],po[1][0],po[1][1],po[2][0],po[2][1]);
88 y=gety(po[0][0],po[0][1],po[1][0],po[1][1],po[2][0],po[2][1]);
89 r=sqrt((po[2][0]-x)*(po[2][0]-x)+(po[2][1]-y)*(po[2][1]-y));
90 }
91 }
92 double temp=sqrt((po[3][0]-x)*(po[3][0]-x)+(po[3][1]-y)*(po[3][1]-y));
93 if(r>temp-1e-8)
94 printf("Case #%d: Danger\n",count);
95 else
96 printf("Case #%d: Safe\n",count);
97 }
98 return 0;
99 }腾讯云开发者
