凸包 点积 Tavas and Pashmaks:CodeForces 535E

题目:CodeForces 535E

time limit per test1 second
memory limit per test256 megabytes
Tavas is a cheerleader in the new sports competition named “Pashmaks”.

This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person’s swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn’t know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn’t really familiar with programming, so he asked you to help him.

Input
The first line of input contains a single integer n (1 ≤ n ≤ 2 × 10^5).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 10^4).

Output
In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples
input
3
1 3
2 2
3 1
output
1 2 3
input
3
1 2
1 1
2 1
output
1 3

题意

有N(N<=2*10^5)名选手参加两场比赛，先游泳后跑步。告诉你每名选手的游泳速度si和跑步速度ri，问哪些选手有可能赢得比赛（泳道和跑道长度未知）。

思路：

假设泳道长度为S，跑道长度为R，则ti=Ssi+Sri。问题就转化为是否存在si和ri使得ti不大于任何的tj。

解法1：

由于1 ≤ si, ri ≤ 10^4，我们可以通过预处理将选手减少到10^4个，然后来个N^2大暴力。

解法2：

题解中的做法是把(ri,si)看成一个个的点，用这些点求凸包，答案就是凸包上从最左到最下（逆时针）的所有点。
为什么呢？
把(R,S)看做向量，每个人所用的时间就是他们所代表的点(ri,si)与这个向量点乘的结果，也就是在这个向量上的投影，那么凸包里面的肯定不如凸包外面的优，右上的肯定不如左下的优。见下图：

对于任意的向量(R,S)，显然与他垂直并且相切在凸包上的点的点积是最小的。所以向量(R,S)在从x轴变化到y轴的过程中。对应的凸包上的切线就是从y轴变化到x轴，这些切线上的点就是凸包的外围点。而且显然靠近坐标原点O的更优。即下图的红线部分。

代码:待补充