在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为
可见的,否则Li为被覆盖的.
例如,对于直线:
L1:y=x; L2:y=-x; L3:y=0
则L1和L2是可见的,L3是被覆盖的.
给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=500000),且n条直线两两不重合.求出所有可见的直线.

　　第一行为N(0 < N < 50000),接下来的N行输入Ai,Bi

　　从小到大输出可见直线的编号，两两中间用空格隔开,最后一个数字后面也必须有个空格

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define MAXN 50010#define Eps 1e-18using namespace std;struct Liyn{  int k, b, pos;  void Push(int i) {scanf("%d%d", &k, &b); pos = i;}  bool operator == (const Liyn &a)const {return k == a.k;}  bool operator < (const Liyn &a)const {return k < a.k || (k == a.k && b > a.b);}  double Cmp(const Liyn &a) {return double(a.b - b) / double(k - a.k);}}L[MAXN], _pb[MAXN];int n, top, ans[MAXN];int main(){  scanf("%d", &n);  for(int i = 0; i < n; i++)    L[i].Push(i);  sort(L, L + n);  n = unique(L, L + n) - L;  for(int i = 0; i < n; i++){    while(top > 1 && _pb[top - 1].Cmp(_pb[top - 2]) > L[i].Cmp(_pb[top - 1]) - Eps)top--;    _pb[top++] = L[i];  }  for(int i = 0; i < top; i++)    ans[i] = _pb[i].pos;  sort(ans, ans + top);  for(int i = 0; i < top; i++)    printf("%d ", ans[i] + 1);  return 0;}