【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;
}