HDU 4720 Naive and Silly Muggles

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 

求三点最小的圆 判断最后一个点是不是在里面

求重心即可

#include<stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
double ll(double a,double b,double c,double d)
{
    return (a-c)*(a-c)+(b-d)*(b-d);
}
int main()
{
    int tes;
    int cas=0;
    double x1,y1,x2,y2,x3,y3,a,b,r,x,y;
    scanf("%d",&tes);
    while(tes--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x,&y);
        a=(x1+x2+x3)/3;
        b=(y1+y2+y3)/3;
        r=99999999;
        r=min(r,ll(a,b,x2,y2));
        r=min(r,ll(a,b,x1,y1));
        r=min(r,ll(a,b,x3,y3));
        if(ll(a,b,x,y)<=r)
            printf("Case #%d: Danger\n",++cas);
        else
            printf("Case #%d: Safe\n",++cas);
    }
    return 0;
}

判断是不是钝角 是的话 最长边的一半就是半径

#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)为钝角
        {
            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)为钝角
        {
            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)为钝角
        {
            a=(x2+x1)/2.0,b=(y2+y1)/2.0;
            r2=(a-x1)*(a-x1)+(b-y1)*(b-y1);
        }
        else
        {
            a=((x1*x1+y1*y1-x2*x2-y2*y2)*(y1-y3)-(x1*x1+y1*y1-x3*x3-y3*y3)*(y1-y2))/(2.0*((y1-y3)*(x1-x2)-(y1-y2)*(x1-x3)));
            b=((x1*x1+y1*y1-x2*x2-y2*y2)*(x1-x3)-(x1*x1+y1*y1-x3*x3-y3*y3)*(x1-x2))/(2.0*((x1-x3)*(y1-y2)-(x1-x2)*(y1-y3)));
            //圆心公式
            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;
}