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

本文链接：https://blog.youkuaiyun.com/ZZ_AC/article/details/49667793

本文介绍了一种通过排序和树状数组来判断多条直线在指定区域内是否有交点的方法，并提供了两种实现思路及其代码示例。

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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 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
   
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.

$\text{[math]}$ 题意：求x1,x2以内的直线是否相交。方法一：判断直线相交，只要直线与x1,x2相交的纵坐标y1,y2和另一直线与x1,x2相交的纵坐标y'1,y'2，满足(y1-y'1)*(y2-y'2)<0 先对直线按y1从小到大排序，y1相等的按照y2从大到小排序。对于一条直线是否与其相交，只要满足 (y1-y'1)*(y2-y'2)<0，找他后面的数(y1-y'1)>0只要找到最小值小于y2的，就可以了对每一个直线都要找，找最小值可以用线段树，树状数组。时间复杂度O(nlongn) 我用的是树状数组。

#include
   
    
#include
    
     
#include
     
      
#include
      
       
#include
       
using namespace std;

const int N = 100001;
long long c[N];

struct node
{
    long long y1;
    long long y2;
}T[N];

bool cm(node x, node y)
{
    if(x.y1 == y.y1)    return x.y2 > y.y2;
    return x.y1 < y.y1;
}

int lowbit(int x)
{
    return x&(-x);
}

void Init(int n)
{
    for(int i = 1; i < n; i++)
    {
        c[i] = T[i].y2;
        for(int j = 1; j < lowbit(i); j <<= 1)
        {
            c[i] = min(c[i], c[i-j]);
        }
    }
}

long long Query(int l, int r)
{
    long long ans = T[r].y2;
    while(true)
    {
        ans = min(ans, T[r].y2);
        if(l == r) break;
        for(r-=1; r-l >= lowbit(r); r -= lowbit(r))
        {
            ans = min(ans, c[r]);
        }
    }
    return ans;
}

int main(void)
{
    int n;
    long long x1, x2;
    scanf("%d", &n);
    scanf("%I64d%I64d", &x1, &x2);
    long long k, b;
    int i;
    int flag = 0;
    for(i = 0; i < n; i++)
    {
        scanf("%I64d%I64d", &k, &b);
        T[i].y1 = k*x1+b;
        T[i].y2 = k*x2+b;
    }
    sort(T, T+n, cm);
    Init(n);
    for(i = 0; i < n-1; i++)
    {
        int j = i;
        while(T[j].y1 == T[i+1].y1) i++;
        if(i+1 > n-1) break;
        long long tmp = Query(i+1, n-1);
        if(T[j].y2 > tmp)
        {
            flag = 1;break;
        }
    }
    if(flag) printf("YES\n");
    else printf("NO\n");
    return 0;
}

方法二：想象一下怎样的直线能够相交，设直线

Line1 ，直线

Line2 与

x=L 的交点分别是

L1,L2 ，与

x=R 的交点分别是

R1,R2 ，设

L1>L2 ，那么如果这两根直线在这个区间相交，则必有

R1<R2 。
所以这个问题就转换成了求逆序对的个数，维护一个树状数组即可。
参考：http://harryguo.me/2015/11/05/Codeforces-329-div2/

#include
    
     
#include
     
      
#include
      
       
#include
       
        
#include
        
using namespace std;

const int N = 100001;
long long c[N];
int a[N];
int b[N];
struct node
{
    long long y1;
    long long y2;
}T[N];

bool cm(node x, node y)
{
    if(x.y1 == y.y1)    return x.y2 < y.y2;
    return x.y1 < y.y1;
}


int main(void)
{
    int n;
    long long x1, x2;
    scanf("%d", &n);
    scanf("%I64d%I64d", &x1, &x2);
    long long k, b;
    int i;
    int flag = 0;
    for(i = 0; i < n; i++)
    {
        scanf("%I64d%I64d", &k, &b);
        T[i].y1 = k*x1+b;
        T[i].y2 = k*x2+b;
    }
    sort(T, T+n, cm);
    for(i = 0; i < n-1; i++)
    {
        if(T[i].y2 > T[i+1].y2)//存在逆序对
        {
            flag = 1;break;
        }
    }
    if(flag) printf("YES\n");
    else printf("NO\n");
    return 0;
}