HDU4720-Naive and Silly Muggles（求三角形外心）

Naive and Silly Muggles

                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                 Total Submission(s): 1098    Accepted Submission(s): 732

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

30 02 01 21 -0.50 02 01 21 -0.60 03 01 11 -1.5

 

Sample Output

Case #1: DangerCase #2: SafeCase #3: Safe

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

 

题意： 给你三个点，找一个最小的圆能把三个点包住。点可以允许在圆内。再给你另一个点，如果该点在圆内或圆上输出Danger，在圆外输出Safe

解题思路：锐角或直角的话最小的圆就是外接圆，钝角的话圆的圆心就是在钝角的对边的中心

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>#include <functional>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;const double exps=1e-8;struct point{    double x,y;};typedef struct point point;struct line{    point a,b;};typedef struct line line;point intersection(line u,line v){    point ret=u.a;    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/             ((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));    ret.x+=(u.b.x-u.a.x)*t;    ret.y+=(u.b.y-u.a.y)*t;    return ret;}double distan(point p1,point p2){    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}point cir(point a,point b,point c){    line u,v;    u.a.x=(a.x+b.x)/2;    u.a.y=(a.y+b.y)/2;    u.b.x=u.a.x-a.y+b.y;    u.b.y=u.a.y-a.x-b.x;    v.a.x=(a.x+c.x)/2;    v.a.y=(a.y+c.y)/2;    v.b.x=v.a.x-a.y+c.y;    v.b.y=v.a.y+a.x-c.x;    return intersection(u,v);}int main(){    int t,cas=1;    point a,b,c,d,e;    scanf("%d",&t);    while(t--)    {        scanf("%lf%lf",&a.x,&a.y);        scanf("%lf%lf",&b.x,&b.y);        scanf("%lf%lf",&c.x,&c.y);        d=cir(a,b,c);        scanf("%lf%lf",&e.x,&e.y);        double ans1=distan(a,d);        double ans2=distan(d,e);        point k;        if(distan(a,b)>distan(b,c)&&distan(a,b)>distan(a,c))        {            k.x=(a.x+b.x)/2;            k.y=(a.y+b.y)/2;            if(ans1>distan(c,k))            {                ans1=distan(c,k);                ans2=distan(e,k);            }        }        else if(distan(b,c)>distan(a,b)&&distan(b,c)>distan(a,c))        {            k.x=(c.x+b.x)/2;            k.y=(c.y+b.y)/2;            if(ans1>distan(a,k))            {                ans1=distan(a,k);                ans2=distan(e,k);            }        }        else        {            k.x=(c.x+a.x)/2;            k.y=(c.y+a.y)/2;            if(ans1>distan(b,k))            {                ans1=distan(b,k);                ans2=distan(e,k);            }        }        if(ans1>=ans2) printf("Case #%d: Danger\n",cas++);        else printf("Case #%d: Safe\n",cas++);    }    return 0;}