【Codeforces Round #299 (Div. 2) E】Tavas and Pashmaks

【链接】 我是链接,点我呀:)

【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[我是一个链接](http://blog.csdn.net/fearlessxjdx/article/details/73327063?readlog)
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 <bits/stdc++.h>
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;
}