Codeforces Round #329 (Div. 2)B. Anton and Lines 贪心

B. Anton and Lines

 

The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are1 ≤ i < j ≤ n and x', y', such that:

• y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
• y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
• x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.

You can't leave Anton in trouble, can you? Write a program that solves the given task.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.

The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.

Output

Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).

Sample test(s)

input

4
1 2
1 2
1 0
0 1
0 2

output

NO

input

2
1 3
1 0
-1 3

output

YES

题意：给你x,y,还有n条直线，问你在x到y的区域内部是否存在交点，不包括x,y上
题解：贪心，我们知道在x,y上必然存在两个交点，所以我们贪心按照左边从小到大排序，如果右边不是递增的则存在交点

///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
inline ll read()
{
    ll x=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//****************************************
#define maxn 1000000+5
#define mod 1000000007

double  k[maxn],b[maxn];

struct ss{

   double l;
   double r;
   int index;
}G[maxn];
int cmp(ss s1,ss s2){
    if(s1.l==s2.l)return s1.r<s2.r;
  else  return s1.l<s2.l;
}
int n;
int main(){

   n=read();
   double x,y;
   scanf("%lf%lf",&x,&y);
   int kk=;
   for(int i=;i<=n;i++){
    scanf("%lf%lf",&k[i],&b[i]);
        double   xx=k[i]*x+b[i];
        double   yy=k[i]*y+b[i];
        G[++kk].l=xx;G[kk].r=yy;
   }
   sort(G+,G+kk+,cmp);
   double now=-,last=G[].r;
   for(int i=;i<=kk;i++){

          if(G[i].r<last){
            cout<<"YES"<<endl;
            return ;

       }
       now=G[i].l;
       last=G[i].r;
   }
   cout<<"NO"<<endl;
  return ;
}

代码