Codeforces Gym 100338 B

题意：给两个线段，要求找一个圆（输出圆心和半径），使得圆和每个线段的内部都有且只有一个交点。。

解法：枚举两个线段两两个点，这样可以得到4对点，找到距离最近的一对，它们的中点就是圆心，距离一半再加上eps就是半径，这题eps = 1e-4。。

Code

//Hello. I‘m Peter.
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;

const double eps = 1e-4;
inline int sgn(double x){
    if(fabs(x) < eps) return 0;
    else return x > 0? 1 : -1;
}

struct Point{
    double x, y;
    Point(){};
    Point(double x1, double y1){
        x = x1, y = y1;
    }
}p[100];
typedef Point Vector;
Vector operator - (Vector a, Vector b){
    return Vector(a.x - b.x, a.y - b.y);
}
double Length(Vector v){
    return sqrt(v.x * v.x + v.y * v.y);
}
Point getp(){
    int x,y;
    scanf("%d%d",&x,&y);
    return Point(x, y);
}

Point ans;
double d;
void update(Point p1, Point p2){
    if(sgn(Length(p1 - p2) - d) < 0){
        d = Length(p1 - p2);
        ans = Point((p1.x + p2.x) * 0.5, (p1.y + p2.y) * 0.5);
    }
}
int main(){
    freopen("geometry.in","r",stdin);
    freopen("geometry.out","w",stdout);
    while(true){
        for(int i = 1; i <= 4; i++) p[i] = getp();
        bool alz = 1;
        for(int i = 1; i <= 4 && alz == 1; i++) if(p[i].x != 0 || p[i].y != 0) alz = 0;
        if(alz) break;

        d = 1e9;
        update(p[1], p[3]);
        update(p[2], p[3]);
        update(p[1], p[4]);
        update(p[2], p[4]);
        d = d * 0.5;
        d = d + 0.001;

        printf("%.10f %.10f %.10f\n",ans.x,ans.y,d);
    }
    return 0;
}