CodeForces 593B - Anton and Lines(思維)

B. Anton and Lines
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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).

Examples
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
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.

題意:
給出n條直線和一個區間,給出斜率和節距來描述直線.
問這些直線在區間中有沒有交點.

解題思路:
對直線落在區間邊界上的y值進行排序,如果都是非嚴格上升(下降),就沒有交點.

AC代碼:

#include<bits/stdc++.h>
using namespace std;
const long long maxn = 1e5+5;
struct node
{
    long long x1,x2;
}a[maxn];
bool cmp(node a,node b)
{
    if(a.x1 < b.x1) return 1;
    else if(a.x1 == b.x1)   return a.x2 < b.x2;
    return 0;
}
int main()
{
    int n;
    ios::sync_with_stdio(false);
    scanf("%d",&n);
    long long l,r;
    scanf("%lld%lld",&l,&r);
    for(long long i = 0;i < n;i++)
    {
        long long k,b;
        scanf("%lld%lld",&k,&b);
        a[i].x1 = k*l+b;
        a[i].x2 = k*r+b;
    }
    sort(a,a+n,cmp);
    for(int i = 1;i < n;i++)
    if(a[i].x2 < a[i-1].x2){cout<<"YES";return 0;}
    cout<<"NO";
    return 0;
}