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

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最新推荐文章于 2023-04-17 01:32:38 发布

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本文介绍了一个几何问题的解决方法，即判断两条直线是否在指定区域内相交。通过计算每条直线与区域边界交点的位置，进而判断是否存在至少两直线在区域内相交的情况。

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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 are 1 ≤ 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

input
2
1 3
1 0
0 2

output
YES

input
2
1 3
1 0
0 3

output
NO

题目大意：

给你两个直线，再给你其他直线的k和b，问是否有在区间内的交点；

解题思路：

每个直线和区间边界的交点，判断在两个交点中海油另一组焦点；

代码：

#include "iostream"
#include "cstdio"
#include "algorithm"
#include "math.h"
#include "string"
#include "string.h"
using namespace std;
struct node
{
    int id; 
    double y1,y2;
}a[100100];

bool cmp(node n1,node n2)
{
    if(n1.y1!=n2.y1)
        return n1.y1>n2.y1;
    return n1.y2>n2.y2;
}

int main(int argc, char* argv[])
{
    int n,i;
    double x1,x2,k,b;
    cin>>n;
    scanf("%lf%lf",&x1,&x2);
    for(i=0;i<n;i++)
    {
        scanf("%lf%lf",&k,&b);
        a[i].id=i;
        a[i].y1=k*x1+b;
        a[i].y2=k*x2+b;
    }
    sort(a,a+n,cmp);
    for(i=1;i<n;i++)
       if(a[i].y2>a[i-1].y2)
            break;
    if(i<n)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
}