• codeforces 593B Anton and Lines


    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 kibi ( - 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
    Note

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

     1 #include<cstdio>
     2 #include<algorithm>
     3 using namespace std;
     4 
     5 struct line
     6 {
     7     long long sy,ey;
     8 }p[100005];
     9 
    10 bool cmp(line a,line b)
    11 {
    12     if(a.sy==b.sy)return a.ey<b.ey;
    13     return a.sy<b.sy;
    14 }
    15 
    16 int main()
    17 {
    18     int n;
    19     while(scanf("%d",&n)!=EOF)
    20     {
    21         int x1,x2;
    22         scanf("%d%d",&x1,&x2);
    23         for(int i=0;i<n;i++)
    24         {
    25             int k,b;
    26             scanf("%d%d",&k,&b);
    27             p[i].sy=(long long)k*x1+b;
    28             p[i].ey=(long long)k*x2+b;
    29         }
    30         sort(p,p+n,cmp);
    31         int k=0;
    32         for(;k<n-1;k++)
    33             if(p[k].sy<p[k+1].sy&&p[k].ey>p[k+1].ey)break;
    34         if(k==n-1)
    35             puts("NO");
    36         else
    37             puts("YES");
    38     }
    39     return 0;
    40 }
  • 相关阅读:
    k8s部署zk集群
    k8s容器探针
    k8s基于Canal来定义网络策略
    k8s StorageClass(NFS)
    k8s pv与pvc
    k8s的弹性伸缩(HPA)
    k8sStatefulSet
    k8s 亲和性
    VUE+SpringBoot环境准备
    oracle获取当前时间,精确到毫秒并指定精确位数的实现方法
  • 原文地址:https://www.cnblogs.com/homura/p/4950882.html
Copyright © 2020-2023  润新知