CGL_2_D: Distance

Distance

For given two segments s1 and s2, print the distance between them.

s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.

Input

The entire input looks like:

q (the number of queries)
1st query
2nd query
...
qth query

Each query consists of integer coordinates of end points of s1 and s2 in the following format:

xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3

Output

For each query, print the distance. The output values should be in a decimal fraction with an error less than 0.00000001.

Constraints

• 1 ≤ q ≤ 1000
• -10000 ≤ xpi, ypi ≤ 10000
• p0≠p1 and p2≠p3.

Sample Input

3
0 0 1 0 0 1 1 1
0 0 1 0 2 1 1 2
-1 0 1 0 0 1 0 -1

Sample Output

1.0000000000
1.4142135624
0.0000000000

---

Source: https://onlinejudge.u-aizu.ac.jp/problems/CGL_2_D

求两个线段的距离

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <cstdio>
// 计算几何模板
//二维平面
using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//Compares a double to zero
int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return-1;
    else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}

struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x = _x;
        y = _y;
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void output(){
        printf("%.2f-%.2f\n",x,y);
    }
    bool operator == (Point b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)== 0-sgn(y-b.y)?0:x<b.x;
    }
    Point operator-(const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    }
    //点积
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);//库函数
    }
    //返回长度的平方
    double len2(){
        return x*x + y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator +(const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator *(const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator /(const double &k)const{
        return Point(x/k,y/k);
    }
    //计算 pa 和 pb 的夹角
    //就是求这个点看 a,b 所成的夹角
    //测试 LightOJ1203
    double rad(Point a,Point b){
        Point p = *this;
        return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
    }
    //化为长度为 r 的向量
    Point trunc(double r){
        double l = len();
        if(!sgn(l))return *this;
        r /= l;
        return Point(x*r,y*r);
    }
    //逆时针旋转 90 度
    Point rotleft(){
        return Point(-y,x);
    }
    //顺时针旋转 90 度
    Point rotright(){
        return Point(y,-x);
    }
    //绕着 p 点逆时针旋转 angle
    Point rotate(Point p,double angle){
        Point v = (*this)-p;
        double c = cos(angle), s = sin(angle);
        return Point(p.x + v.x*c-v.y*s,p.y + v.x*s + v.y*c);
    }
};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s = _s;
        e = _e;
    }
    bool operator ==(Line v){
        return (s == v.s)&&(e == v.e);
    }
    //根据一个点和倾斜角 angle 确定直线,0<=angle<pi
    Line(Point p,double angle){
        s = p;
        if(sgn(angle-pi/2) == 0){
            e = (s + Point(0,1));
        }
        else{
            e = (s + Point(1,tan(angle)));
        }
    }
    //ax+by+c=0
    Line(double a,double b,double c){
        if(sgn(a) == 0){
            s = Point(0,-c/b);
            e = Point(1,-c/b);
        }
        else if(sgn(b) == 0){
            s = Point(-c/a,0);
            e = Point(-c/a,1);
        }
        else{
            s = Point(0,-c/b);
            e = Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e < s){
            swap(s,e);
        }
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //返回直线倾斜角 0<=angle<pi
    double angle(){
        double k = atan2(e.y-s.y,e.x-s.x);
        if(sgn(k) < 0)k += pi;
        if(sgn(k-pi) == 0)k-= pi;
        return k;
    }
    //点和直线关系
    //1 在左侧
    //2 在右侧
    //3 在直线上
    int relation(Point p){
        int c = sgn((p-s)^(e-s));
        if(c < 0)return 1;
        else if(c > 0)return 2;
        else return 3;
    }
    // 点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
    }
    //两向量平行 (对应直线平行或重合)
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s)) == 0;
    }
    //两线段相交判断
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int segcrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        int d3 = sgn((v.e-v.s)^(s-v.s));
        int d4 = sgn((v.e-v.s)^(e-v.s));
        if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
        return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
               (d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
               (d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
               (d4==0 && sgn((e-v.s)*(e-v.e))<=0);
    }
    //直线和线段相交判断
    //-*this line -v seg
    //2 规范相交
    //1 非规范相交
    //0 不相交
    int linecrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //两直线关系
    //0 平行
    //1 重合
    //2 相交
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //求两直线的交点
    //要保证两直线不平行或重合
    Point crosspoint(Line v){
        double a1 = (v.e-v.s)^(s-v.s);
        double a2 = (v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1
        ));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
            return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
    //返回线段到线段的距离
    //前提是两线段不相交，相交距离就是 0 了
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
    //返回点 p 在直线上的投影
    Point lineprog(Point p){
        return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
    }
    //返回点 p 关于直线的对称点
    Point symmetrypoint(Point p){
        Point q = lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};


int main(){
    int t;
    cin>>t;
    while(t--){
        Line a,b;
        cin>>a.s.x>>a.s.y>>a.e.x>>a.e.y>>b.s.x>>b.s.y>>b.e.x>>b.e.y;
        double ans=0.0;
        if(a.segcrossseg(b)==0) ans=a.dissegtoseg(b);
        cout<<fixed<<setprecision(10)<<ans<<"\n";
    }
    return 0;
}