HDU 4720 Naive and Silly Muggles (外切圆心)
Naive and Silly Muggles
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 228 Accepted Submission(s): 163
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?
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.
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
Case #2: Safe
Case #3: Safe
外切圆心坐标
x=(x1+x2+x3)/3;
y=(y1+y2+y3)/3;
import java.awt.Point;
import java.io.*;
import java.util.*; public class Main {
BufferedReader bu;
PrintWriter pw;
int t;
double x,y;
public static void main(String[] args) throws Exception {
new Main().work();
} void work() throws Exception {
Scanner sc=new Scanner(new InputStreamReader(System.in));
pw = new PrintWriter(new OutputStreamWriter(System.out), true);
t = sc.nextInt();
for (int p = 1; p <= t; p++) {
pw.print("Case #" + p + ": ");
double x1, y1;
double x2, y2;
double x3, y3;
double x4, y4;
//第一个wizard 的坐标
x1 = sc.nextDouble();
y1 = sc.nextDouble();
//第二个wizard 的坐标
x2 = sc.nextDouble();
y2 = sc.nextDouble();
//第三个wizard 的坐标
x3 = sc.nextDouble();
y3 = sc.nextDouble();
//muggles 坐标
x4 = sc.nextDouble();
y4 = sc.nextDouble();
//外切园的圆心坐标
x=(x1+x2+x3)/3;
y=(y1+y2+y3)/3;
//半径
double r=Math.sqrt((x1-x)*(x1-x)+(y1-y)*(y1-y));
//muggles到圆心的距离
double d=Math.sqrt((x4-x)*(x4-x)+(y4-y)*(y4-y)); if(d>r){
pw.println("Safe");
}
else{
pw.println("Danger");
}
}
}
}
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