hdu4720

Naive and Silly Muggles

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 216    Accepted Submission(s): 153

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

 

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

 

一定要考虑钝角三角形和锐角三角形的区别，利用向量法判断三角形中是否有钝角

 

 

#include<iostream>
#include<cstring>
#include<cmath>
#include<string>
#include<cstdio>
using namespace std;

int main()
{
    int tes;
    int cas=0;
    double x1,y1,x2,y2,x3,y3,a,b,r2,x,y;
    scanf("%d",&tes);
    while(tes--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x,&y);
        if((x2-x1)*(x3-x1)+(y2-y1)*(y3-y1)<0)            //（x1，y1）是钝角
        {
            a=(x3+x2)/2.0,b=(y3+y2)/2.0;
            r2=(a-x2)*(a-x2)+(b-y2)*(b-y2);
        }
        else if((x1-x2)*(x3-x2)+(y1-y2)*(y3-y2)<0)        //（x2，y2）是钝角
        {
            a=(x3+x1)/2.0,b=(y3+y1)/2.0;
            r2=(a-x1)*(a-x1)+(b-y1)*(b-y1);
        }
        else if((x1-x3)*(x2-x3)+(y1-y3)*(y2-y3)<0)        //（x3，y3）是钝角
        {
            a=(x2+x1)/2.0,b=(y2+y1)/2.0;
            r2=(a-x1)*(a-x1)+(b-y1)*(b-y1);
        }
        else         //三角形是锐角三角形
        {
            a=((y2-y1)*(y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1))/(2.0*((x3-x1)*(y2-y1)-(x2-x1)*(y3-y1)));
            b=((x2-x1)*(x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1))/(2.0*((y3-y1)*(x2-x1)-(y2-y1)*(x3-x1)));
            r2=(x1-a)*(x1-a)+(y1-b)*(y1-b);
        }
        if((x-a)*(x-a)+(y-b)*(y-b)<=r2)
            printf("Case #%d: Danger\n",++cas);
        else
            printf("Case #%d: Safe\n",++cas);
    }
    return 0;
}