【链接】

【题意】

游泳+跑步比赛。

先游泳,然后跑步.

最先到终点的人是winner.

但是现在游泳的距离和跑步的距离长度都不确定。

S和R.

给你n个人,告诉你每个人游泳的速度以及它们跑步的速度。

现在问你在改变S,R的情况下,第i个人有没有可能为winner.

输出所有可能为winner的人的编号。

【题解】

每个人所花费的时间为\(\frac{S}{si} + \frac{R}{ri}\)

可以看成是向量{S,R}和(\(\frac{1}{si}\),\(\frac{1}{ri}\))的点积。

考虑点积的几何意义为向量上的投影长度。

而一堆点里面,和某个固定的向量的投影长度.↓↓

会发现,总是凸包的边界上的某个点。

接下来的思路来自这里

看完思路之后接着

则,只要求出凸包。

然后找最坐标的x值最小的点就好(有多个x相同就找y尽量小的);

->求完凸包之后,因为排了个序,ch[0]就是这个最左的点。

然后找一个y值为最小的点就好了。

(多个最小,直接取x最小的就好);

所以求完凸包之后,顺序枚举,找到一个y=y的最小值,那么这一段就是所需的了。

重复点不要忘记加上去~就是那个to.

【代码】

#include 
    
     using namespace std;struct Point {    double x, y;    int id;    Point() {}    Point(double x, double y) {        this->x = x;        this->y = y;    }    void read(int id) {        int s, b;        scanf("%d%d", &s, &b);        x = 1000000.0 / s; y = 1000000.0 / b;        this->id = id;    }};const double eps = 1e-16;int dcmp(double x) {    if (fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}typedef Point Vector;Vector operator + (Vector A, Vector B) {    return Vector(A.x + B.x, A.y + B.y);}Vector operator - (Vector A, Vector B) {    return Vector(A.x - B.x, A.y - B.y);}Vector operator * (Vector A, double p) {    return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {    return Vector(A.x / p, A.y / p);}bool operator < (const Point& a, const Point& b) {    return a.x < b.x || (dcmp(a.x - b.x) == 0 && a.y < b.y);}bool operator == (const Point& a, const Point& b) {    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积struct Line {    Point v, p;    Line() {}    Line(Point v, Point p) {        this->v = v;        this->p = p;    }    Point point(double t) {        return v + p * t;    }};//向量旋转Vector Rotate(Vector A, double rad) {    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));}Vector AngleBisector(Point p, Vector v1, Vector v2){//给定两个向量，求角平分线    double rad = Angle(v1, v2);    return Rotate(v1, dcmp(Cross(v1, v2)) * 0.5 * rad);}//判断3点共线bool LineCoincide(Point p1, Point p2, Point p3) {    return dcmp(Cross(p2 - p1, p3 - p1)) == 0;}//判断向量平行bool LineParallel(Vector v, Vector w) {    return Cross(v, w) == 0;}//判断向量垂直bool LineVertical(Vector v, Vector w) {    return Dot(v, w) == 0;}//计算两直线交点，平行，重合要先判断Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {    Vector u = P - Q;    double t = Cross(w, u) / Cross(v, w);    return P + v * t;}//点到直线距离double DistanceToLine(Point P, Point A, Point B) {    Vector v1 = B - A, v2 = P - A;    return fabs(Cross(v1, v2)) / Length(v1);}//点到线段距离double DistanceToSegment(Point P, Point A, Point B) {    if (A == B) return Length(P - A);    Vector v1 = B - A, v2 = P - A, v3 = P - B;    if (dcmp(Dot(v1, v2)) < 0) return Length(v2);    else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);    else return fabs(Cross(v1, v2)) / Length(v1);}//点在直线上的投影点Point GetLineProjection(Point P, Point A, Point B) {    Vector v = B - A;    return A + v * (Dot(v, P - A) / Dot(v, v));}//线段相交判定(规范相交)bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}//可以不规范相交bool SegmentProperIntersection2(Point a1, Point a2, Point b1, Point b2) {      double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),              c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);      return max(a1.x, a2.x) >= min(b1.x, b2.x) &&      max(b1.x, b2.x) >= min(a1.x, a2.x) &&      max(a1.y, a2.y) >= min(b1.y, b2.y) &&      max(b1.y, b2.y) >= min(a1.y, a2.y) &&      dcmp(c1) * dcmp(c2) <= 0 && dcmp(c3) * dcmp(c4) <= 0;  }//判断点在线段上, 不包含端点bool OnSegment(Point p, Point a1, Point a2) {    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;}//n边形的面积double PolygonArea(Point *p, int n) {    double area = 0;    for (int i = 1; i < n - 1; i++)        area += Cross(p[i] - p[0], p[i + 1] - p[0]);    return area / 2;}//点是否在多边形内部int isPointInPolygon(Point o, Point *p, int n) {    int wn = 0;    for (int i = 0; i < n; i++) {        if (OnSegment(o, p[i], p[(i + 1) % n])) return -1;        int k = dcmp(Cross(p[(i + 1) % n] - p[i], o - p[i]));        int d1 = dcmp(p[i].y - o.y);        int d2 = dcmp(p[(i + 1) % n].y - o.y);        if (k > 0 && d1 <= 0 && d2 > 0) wn++;        if (k < 0 && d2 <= 0 && d1 > 0) wn--;    }    if (wn != 0) return 1;    return 0;}const int N = 200005;int to[N];//凸包int ConvexHull(Point *p, int n, Point *ch) {    sort(p, p + n);    int m = 0;    memset(to, -1, sizeof(to));    //两个叉积改成<,可以求共线凸包    for (int i = 0; i < n; i++) {        to[i] = i;        if (i && p[i] == p[i - 1]) {            to[i] = to[i - 1];            continue;        }        while (m > 1 && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;        ch[m++] = p[i];    }    int k = m;    for (int i = n - 2; i >= 0; i--) {        while (m > k && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;        ch[m++] = p[i];    }    if (n > 1) m--;    return m;}Point a[N+10],ch[N+10];bool vis[N+10];int n,m;double mi = 1e18;int main(){    #ifdef LOCAL_DEFINE        freopen("F:\\c++source\\rush_in.txt","r",stdin);                #endif    scanf("%d",&n);    for (int i = 0;i < n;i++){        a[i].read(i);    }    m = ConvexHull(a,n,ch);    for (int i = 0;i < m;i++) mi = min(mi,ch[i].y);     for (int i = 0;i < m;i++){        vis[ch[i].id] = 1;        if (ch[i].y==mi) break;    }    for (int i = 0;i < n;i++)        if (vis[a[to[i]].id]) vis[a[i].id] = 1;    for (int i = 0;i < n;i++)        if (vis[i]){            printf("%d",i+1);            if (i==n-1) puts("");else putchar(' ');        }    return 0;}