hdu4637 calculation time interleaving two moving object

Title: http: //acm.hdu.edu.cn/showproblem.php pid = 4637?

Reference from: https: //www.cnblogs.com/javawebsoa/p/3239001.html

#include<bits/stdc++.h>
using namespace std;
#define pii pair <double, double>
#define mp make_pair
#define pb push_back
#define X first
#define Y second
const double eps = 1e-8;
int dcmp(double x) {
    if (fabs(x) < eps)
        return 0;
    return x > eps ? 1 : -1;
}
struct point {
    double x, y;
    point() {

    }
    point(double x, double y) :
            x (x) and (y) {
    }
    double operator *(const point &t) const {
        return x * t.x + y * t.y;
    }
    point operator -(const point &t) const {
        return point(x - t.x, y - t.y);
    }
    point operator +(const point &t) const {
        return point(x + t.x, y + t.y);
    }
    point operator *(const double &t) const {
        return point(t * x, t * y);
    }
} He is, and;

double v1, v2, v, t, x, T;
double ans;
int n;
inline double F(double x) {
    return x * x;
}
double cross(const point &o, const point &a, const point &b) {
    return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
double dis(const point &a, const point &b) {
    return sqrt(F(a.x - b.x) + F(a.y - b.y));
}

BOOL segSegIntersect ( const Point & A, const Point & B, const Point & L, const Point & R & lt) { // two line segments intersect (irrespective collinear) 
    return Cross (A, B, L) * Cross (A, B, R & lt) < EPS
             && Cross (L, R & lt, A) * Cross (L, R & lt, B) < EPS;
}

double intersect(const point &a, const point &b, const point &l, const point &r) {//俩直线求交点的x
    double ret = a.x;
    double t = ((a.x - l.x) * (l.y - r.y) - (a.y - l.y) * (l.x - r.x))
            / ((a.x - b.x) * (l.y - r.y) - (a.y - b.y) * (l.x - r.x));
    return ret + (b.x - a.x) * t;
}

Vector < Double > VEC; // record the intersection with raindrops 
Vector <PII> RES;    // record is hit each time segment raindrops 
struct Rain {
    point o, a, b, c;
    double r, h;
    void in() {
        CIN >> OX Oy >> >> >> R & lt H;
         /// record only triangular raindrops to 
        A = O, B = O, C = O;
        a.x -= r;
        b.x += r;
        c.y += h;
    }
    BOOL Inside ( const Point & P) {       // point are raindrops inside (including the boundary) 
        return (DIS (O, P) - EPS <&& R & lt Py - EPS < Oy)
                 || (Cross (C, A, P)> Cross && -EPS (C, B, P) < EPS
                         && Py> Oy + EPS);
    }
    void Query1 () { // semicircular cross raindrops find the intersections 
        Point STA = B, D = ED- STA;
         Double A = D * D;
         Double B = (BO) * * D 2 ;
         Double C = (BO) * (BO) -R & lt * R & lt;
         Double dlt * = the B-B . 4 * a * C;
         IF (dlt <-EPS) return ;
         IF (dlt <EPS) dlt = 0 ;         // eliminate the negative case dlt zero 
        else = DLT sqrt (DLT);

        double t=(-B-dlt)/(2*A);
        TP Point = D + B * T;
         /// semicircular one-two possible points of intersection into a 
        IF (EPS-tp.x <sta.x && tp.x + EPS> EPS-ed.x && tp.y <Oy)     // because it is semi-circular, pay attention to the useless point out the sentence 
            vec.pb (tp.x);

        t=(-B+dlt)/(2*A);
        tp=b+d*t;
        if (tp.x-eps<sta.x&&tp.x+eps>ed.x&&tp.y-eps<o.y)
            vec.pb(tp.x);
    }
    void of query2 () { // triangular cross raindrops find the intersections (not horizontal line segment therein, and since only counted out, does not affect the horizontal) 
        Double X;
         IF (segSegIntersect (A, C, STA, ED)) {
            x=intersect(a,c,sta,ed);
            if(x-eps>ed.x&&x+eps<sta.x)
                vec.pb(x);
        }
        if(segSegIntersect(c,b,sta,ed)){
            x=intersect(c,b,sta,ed);
            if(x-eps>ed.x&&x+eps<sta.x)
                vec.pb(x);
        }
    }
    void solve() {
        vec.clear();
        query1();
        query2();
        /// determining start and end points have not within the scope of the current point 
        IF (Inside (STA)) vec.pb (sta.x);
         IF (Inside (ED)) vec.pb (ed.x);
        sort (vec.begin (), vec.end ());
        int m = UNIQUE (vec.begin (), vec.end ()) - vec.begin ();
         IF (m> = 2 ) // taking the maximum and minimum of the two points of intersection is to be hit raindrops period end points 
            res.pb (MP (VEC [ 0 ], VEC [M- . 1 ]));
    }
}on one;
void init(){
    
}
int main () {
     int   event;
    cin>>cas;
    for (int p=1;p<=cas;p++){
        cin>>v1>>v2>>v>>t>>x>>n;
        T=v1*t/(v2-v1)+t;
        sta.x = x;
        sta.y = 0 ;
        ed.x=x-v1*T;
        ed.y=v*T;
        years = 0 ;
        res.clear();
        for(int i=0;i<n;i++){
            num.in();
            num.solve();
        }
        sort(res.begin(), res.end());
        double r=ed.x;
        for (int i=0;i<res.size();i++){
            if(res[i].X-eps<r&&r-eps<res[i].Y){
                ans+=res[i].Y-r;
                r=res[i].Y;
            }
            else if(r-eps<res[i].X){
                years + nothing = [i] .Y- nothing [i] .x;
                r=res[i].Y;
            } 
        }
        printf("Case %d: %.4f\n",p,ans/v1);
    }
    return 0;
}

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