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

B. Anton and Lines

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/593/problem/B

Description

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 = k_i·x + b_i. 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' = k_i * x' + b_i, that is, point (x', y') belongs to the line number i;
y' = k_j * x' + b_j, 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 k_i, b_i ( - 1 000 000 ≤ k_i, b_i ≤ 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 k_i ≠ k_j, or b_i ≠ b_j.

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 Input

Sample Output

NO

HINT

题意

给你一堆直线，然后问你在(x1,x2)区间内，是否有交点

题解：

这道题转化为，在直线交x = x1,x = x2之后，是否存在逆序对

有两个坑点，1.答案会爆long long 2.区间是开区间

代码

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;

struct node
{
    double x;
    int y;
};
bool cmp(node a,node b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
vector<node> Q;
vector<node> P;
int main()
{
    int n;scanf("%d",&n);
    double x1,x2;
    cin>>x1>>x2;
    if(x1>x2)swap(x1,x2);
    x1 += 1e-8;
    x2 -= 1e-8;
    for(int i=1;i<=n;i++)
    {
        double k,b;
        scanf("%lf%lf",&k,&b);
        node ppp;
        ppp.x = k*x1+b,ppp.y=i;
        Q.push_back(ppp);
        ppp.x = k*x2+b,ppp.y=i;
        P.push_back(ppp);
    }
    sort(Q.begin(),Q.end(),cmp);
    sort(P.begin(),P.end(),cmp);
    int flag = 0;
    for(int i=0;i<n;i++)
    {
        if(Q[i].y!=P[i].y)
        {
            puts("YES");
            return 0;
        }
    }
    puts("NO");
}