[HNOI2008] Horizontal visible Line

Description

There are n linear l1,l2 on the Xoy Cartesian plane,... Ln, if the Y value is positive infinity down, you can see a sub-segment of Li, it is said that Li is visible, otherwise li is covered.
For example, for a line:
L1:y=x; L2:y=-x; L3:y=0
Then L1 and L2 are visible, and L3 are covered.
gives a line of N, expressed as a form of y=ax+b (| a|,| b|<=500000), and the N line 22 does not coincide. Find all the visible lines.

Input

The first behavior n (0 < n < 50000), the next n-line input Ai,bi

Output

From small to large output visible line number, 22 is separated by a space, the last number must also have a space behind

Sample Input3
-1 0
1 0
0 0Sample Output1 2Problem Solving Ideas:This is a geometric water problem, the algorithm is very simple, first according to the slope sort, sequentially processing each straight line y-kx1-b1=0 Y-kx2-b2=0 obtained two linear focal axis, we will find that the convex hull composed of straight line with the K value increment, the intersection of the left side is also incremented so with a monotone stack to maintain the line set,  If the intersection (K,J) is to the left of the intersection (I,J), the line I must be overwritten so the line I pops up the stack, and so on. AC Code:

#include <cstdio>
#include <stack>
#include <algorithm>
#define EPS 1e-8
#define MAXN 50001
using namespace Std;
typedef double D;
Class node{
Public
int num;
D k,b;
};
stack<node>q;
Node T[MAXN],H[MAXN];
BOOL CMP (const node A,CONST node B) {
Return (A.K&LT;B.K) | | ((a.k-b.k<eps) && (a.b>b.b)); Problem 1
}
BOOL Cmp2 (const node A,const node B) {
Return a.num<b.num;
}
Double Public_node (node A,node B) {
Return (B.B-A.B)/(A.K-B.K);
}
BOOL Istrue (int i) {
Node tmp=q.top (), TMP2;
Q.pop ();
if (! Q.empty ())
Tmp2=q.top ();
Else
tmp2=tmp;
Q.push (TMP);
if (Public_node (H[I],TMP2) <=public_node (TMP,TMP2))
return true;
return false;
}
int main () {
int n;
scanf ("%d", &n);
for (int i=0;i<n;++i)
scanf ("%lf%lf", &h[i].k,&h[i].b),
H[i].num=i;
Sort (h,h+n,cmp);
Q.push (H[0]);
for (int i=1;i<n;++i) {
if ((H[I].K-H[I-1].K) <eps)//Problem 2
Continue
while (Q.size () >1&&! Q.empty () &&istrue (i))
Q.pop ();
Q.push (H[i]);
}
int i=0;
while (! Q.empty ()) {
T[i++]=q.top ();
Q.pop ();
}
Sort (T,T+I,CMP2);
for (int j=0;j<i;++j)
printf ("%d", t[j].num+1);
}

Note the situation:Note in the code P1 P2, remember not to write the k,b in Node as shaping, if the write as a floating-point shape can not be used on the logical relationship is equal toCommonsense Knowledge: Two floating-point numbers can only be compared in a bad way.

[HNOI2008] Horizontal visible Line