foj 2144 三位几何+区间覆盖

题目大意：一个人站在三维坐标系下的原点处用炮打蚊子，给出n个蚊子的起始坐标跟单位时间匀速移动的方向向量，距离他R以内的蚊子都可以打到，不过他也需要休息，没蚊子的时候也可以休息下。求他要起来多少次打蚊子。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = ;
const double eps = 1e-;
const double Pi=acos(-1.0);

int n, cnt;
double R;

inline bool scan_d(double &ret)
{  //如果数据比较多，而输入的数值比较小就不用scanf而用这种输入方法，用scanf会超时
    char c; int sgn;
    if(c = getchar(),c == EOF) return ; //EOF
    while(c != '-' && (c < '' || c > '')) c = getchar();
    sgn = (c == '-') ? - : ;
    ret = (c == '-') ?  : (c - '');
    while(c=getchar(),c>=''&&c<='') ret=ret*+(c-'');
    ret *= sgn;
    return ;
} 

struct Point3
{
    double x, y, z;
    Point3(double x=,double y=,double z=):x(x),y(y),z(z){}
    void get() { scan_d(x); scan_d(y); scan_d(z); }
};

struct Node
{
    double x, y;
}f[N];

double Dot(Point3 a,Point3 b){return a.x*b.x+a.y*b.y+a.z*b.z;}

inline double min(double a,double b){ return a-b>eps?b:a;}
inline double max(double a,double b){ return a-b>eps?a:b;}

void Deal(Point3 p, Point3 q)
{
    double a=Dot(q,q);
    double b=*Dot(p,q);
    double c =Dot(p,p)-R*R;
    double t =b*b-*a*c;
    if (t < eps) return;
    else if (fabs(t) < eps)
    {
        double i = -b / (2.0 * a);
        if (i < ) return;
        f[cnt].x = f[cnt].y = i;
    }
    else
    {
        double i = (-b + sqrt(t) ) / ( * a);
        double j = (-b - sqrt(t) ) / ( * a);
        if (i < -eps && j < -eps) return;
        else if (i < -eps) i = ;
        else if (j < -eps) j = ;
        f[cnt].x = min(i, j);
        f[cnt].y = max(i, j);
    }
    cnt++;
} 

void Init()
{
    cnt = ;
    Point3 p, v;
    scanf("%d%lf", &n, &R);
    for (int i = ; i < n; i++)
    {
        p.get();v.get();
        Deal(p, v);
    }
}

bool mycomp(const Node &a, const Node &b)
{
    if (fabs(a.y - b.y) > eps) return a.y - b.y < eps;
    return a.x - b.x < eps;
}

int Solve()
{
    int i,ans = ;
    double start = -;
    for (i = ; i < cnt; i++)
    {
        if (f[i].x - start > eps)
        {
            start = f[i].y;
            ans++;
        }
    }
    return ans;
}

int main ()
{
    int i,Icas;
    scanf("%d", &Icas);
    for (i = ; i <= Icas; i++)
    {
        Init();
        sort(f, f + cnt, mycomp);
        printf("Case %d: %d %d\n", i, cnt, Solve());
    }
    return ;
}