hdu4637 计算俩运动对象的时间交

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

参考自：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), y(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);
    }
}sta,ed;

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) {//两线段相交(不考虑共线)
    return cross(a, b, l) * cross(a, b, r) < eps
            && cross(l, r, a) * cross(l, r, 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; //记录与雨滴的交点
vector<pii>res;   //记录被雨滴打到的每个时间段
struct rain {
    point o, a, b, c;
    double r, h;
    void in() {
        cin>>o.x>>o.y>>r>>h;
        ///只记录雨滴的三角形即可 
        a = o, b = o, c = o;
        a.x -= r;
        b.x += r;
        c.y += h;
    }
    bool inside(const point &p) {      //点是否在雨滴里面（包括边界）
        return (dis(o, p) - eps < r && p.y - eps < o.y)
                || (cross(c, a, p) > -eps && cross(c, b, p) < eps
                        && p.y > o.y + eps);
    }
    void query1(){//与雨滴的半圆 交 求交点
        point b=sta,d=ed-sta;
        double A=d*d;
        double B=(b-o)*d*2;
        double C=(b-o)*(b-o)-r*r;
        double dlt=B*B-4*A*C;
        if(dlt<-eps)return;
        if(dlt<eps)dlt=0;        //消除dlt负数零的情况
        else dlt=sqrt(dlt);

        double t=(-B-dlt)/(2*A);
        point tp=b+d*t;
        ///半圆可能的俩个一进一出交点 
        if (tp.x-eps<sta.x&&tp.x+eps>ed.x&&tp.y-eps<o.y)    //因为是半圆，注意把没用的点判掉
            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 query2(){ //与雨滴的三角形 交 求交点 （水平的线段不算在其中,因为只算进和出，水平线不影响）
        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();
        ///判断起点终点有没有在当前点的范围内 
        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)//取最大和最小的两个交点就是被雨滴打到的时间段的两端点
            res.pb(mp(vec[0],vec[m-1]));
    }
}num;
void init(){
    
}
int main() {
    int  cas;
    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;
        ans=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){
                ans+=res[i].Y-res[i].X;
                r=res[i].Y;
            } 
        }
        printf("Case %d: %.4f\n",p,ans/v1);
    }
    return 0;
}