Codeforces Round #299 (Div. 1)C. Tavas and Pashmaks （凸壳）

C. Tavas and Pashmaks

 

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 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

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

Sample test(s)

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：n个人比赛，游泳和赛跑，游泳距离S，赛跑R。 每个人对应两个速度（陆地和水上的）， 如果存在 S ， R，使得第i个人胜利，那么输出i。
题目要求 ：输出所有的i。

可以把 ri si的倒数看成坐标系上的点，时间可以当做 两个向量的点积也就是投影。。。
 求出凸包。 p1 为最下面的点如果有多个选取最左面的那个， p2位最左面的点如果有多个选取最下面的那个。 那么凸包上从p1到p2的点必然满足题意。注意判重

 #include <bits/stdc++.h>
 using namespace std;
 const double eps = 1e-;
 const int inf = 0x3f3f3f3f;
 struct Point
 {
     double x, y;
     int idx;
     Point (double x = , double y = ):
         x(x), y(y) {}
     bool operator == (const Point &rhs)const
     {
         return abs(x - rhs.x) < eps && (y - rhs.y) < eps;
     }
     bool operator < (const Point &rhs) const
     {
         return x < rhs.x || (abs(x-rhs.x) < eps && y < rhs. y);
     }
 };
 typedef Point Vector;
 Vector operator - (Point p1, Point p2)
 {
     return Vector (p1.x-p2.x, p1.y-p2.y);
 }
 double Cross(Vector p1, Vector p2)
 {
     return p1.x*p2.y - p2.x*p1.y;
 }
 Point p[], cvx[];
 bool ans[];
 int pos[];
 bool cmp(double x)
 {
     return x <  || abs(x) < eps;
 }
 int ConvexHull(int n)
 {
     sort (p, p+n);
     // n = unique(p, p+n) - p;
     int tot = ;
     for (int i = ; i < n; i++)
     {
         if (i >  && p[i] == p[i-])
         {
             pos[i] = pos[i-];
             continue;
         }
         pos[i] = i;
         while (tot > && cmp(Cross(cvx[tot-]-cvx[tot-], p[i]-cvx[tot-])) == true)
             tot--;
         cvx[tot++] = p[i];
     }
     int k = tot;
     for (int i = n-; i >= ; i--)
     {
         while (tot > k && cmp(Cross(cvx[tot-]-cvx[tot-],p[i]-cvx[tot-]) == true))
             tot--;
         cvx[tot++] = p[i];
     }
     if (n > )
         tot--;
     return tot;
 }
 bool cmp2(const Point &p1, const Point &p2)
 {
     return p1.y < p2.y || (abs(p1.y-p2.y) < eps && p1.x < p2.x);
 }
 int main(void)
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
 #endif // ONLINE_JUDGE
     int n;
     while (~scanf ("%d", &n))
     {
         memset(ans, false, sizeof(ans));
         memset(pos, , sizeof(pos));
         double minv1 = inf, minv2 = inf;
         for (int i = ; i < n; i++)
         {
             double s, r;
             scanf ("%lf%lf", &s, &r);
             minv1 = min(/r, minv1);    //减小误差
             minv2 = min(minv2, /s);
             p[i] = Point(/s, /r);
             p[i].idx = i;
         }
         int tot = ConvexHull(n);
         for  (int i = ; i < tot; i++)
         {
             ans[cvx[i].idx] = true;
             if (abs(cvx[i].y-minv1) < eps)
                 break;
         }
         for (int i = ; i < n; i++)
         {
             if (ans[p[pos[i]].idx] == true)
                 ans[p[i].idx] = true;
         }
         for (int i = ; i < n; i++)
             if (ans[i] == true)
                 printf("%d ", i+);
         printf("\n");

     }

     return ;
 }