Codeforces 593B Anton and Lines 【思维】

B. Anton and Lines

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 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

41 21 21 00 10 2

output

NO

input

21 31 0-1 3

output

YES

input

21 31 00 2

output

YES

input

21 31 00 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.

题意：给你n条直线，问你在[x1, x2]的开区间里面存不存在至少一个交点。

思路：根据直线的k 和 b求出与x = x1 和 x = x2的交点坐标，分别用lv和rv来记录。

这样只要存在i和j(i != j)满足num[i].lv > num[j].lv && num[i].rv < num[j].rv就可以判定存在交点。

关键在于如何高效的查找。我们可以先按lv升序排列，再按rv升序排列，然后依次比较。

分析比较过程：(数组下标从1-n)首先取一个k = 1，然后遍历i(2 <= i <= n).

我们来比较num[k]和num[i]的lv、rv

(1) 满足上面的判定条件，存在交点；

(2) num[k].lv == num[i].lv，为了最大概率找到交点，我们取rv值较小的代替k向下比较，这样k = i；

(3) num[k].rv > num[i].rv，同(2)，k = i；

(4) num[k].rv == num[i].rv，同(2)， k = i；

上面的后三种处理可以放在一起。这样查询只需要O(n)的时间复杂度。

注意：lv和rv会超int，一开始还以为自己思路错了。。。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN 100000+10#define MAXM 50000000#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;struct Node{    LL lv, rv;};Node num[MAXN];bool cmp(Node a, Node b){    if(a.lv != b.lv)        return a.lv > b.lv;    else        return a.rv > b.rv;}int main(){    int n; Ri(n);    LL x1, x2;    Rl(x1); Rl(x2);    for(int i = 1; i <= n; i++)    {        LL k, b;        Rl(k); Rl(b);        num[i].lv = k * x1 + b;        num[i].rv = k * x2 + b;    }    sort(num+1, num+n+1, cmp);    bool flag = false;    int k = 1;    for(int i = 2; i <= n; i++)    {        if(num[k].lv > num[i].lv && num[k].rv < num[i].rv)        {            flag = true;            break;        }        else            k = i;    }    if(flag)        printf("YES\n");    else        printf("NO\n");    return 0;}