Educational Codeforces Round 46 (Rated for Div. 2) C - Covered Points Count_you are given n n segments on a coordinate line; e

作者：一键难忘520 | 2024-08-01 06:06:19

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you are given n n segments on a coordinate line; each endpoint of every segm

You are given $n$ segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every $k \in [1.. n]$ , calculate the number of points with integer coordinates such that the number of segments that cover these points equals $k$ . A segment with endpoints $l_{i}$ and $r_{i}$ covers point $x$ if and only if $l_{i} \leq x \leq r_{i}$ .

Input

The first line of the input contains one integer $n$ ( $1 \leq n \leq 2 \cdot 10^{5}$ ) — the number of segments.

The next $n$ lines contain segments. The $i$ -th line contains a pair of integers $l_{i}, r_{i}$ ( $0 \leq l_{i} \leq r_{i} \leq 10^{18}$ ) — the endpoints of the $i$ -th segment.

Output

Print $n$ space separated integers $c n t_{1}, c n t_{2}, \dots, c n t_{n}$ , where $c n t_{i}$ is equal to the number of points such that the number of segments that cover these points equals to $i$ .

   Examples 
 

     input 
    
      Copy 
    
3
0 3
1 3
3 8

     output 
    
      Copy 
    
6 2 1 

     input 
    
      Copy 
    
3
1 3
2 4
5 7

     output 
    
      Copy 
    
5 2 0 

Note

The picture describing the first example:

$\text{[math]}$

Points with coordinates $[0, 4, 5, 6, 7, 8]$ are covered by one segment, points $[1, 2]$ are covered by two segments and point $[3]$ is covered by three segments.

The picture describing the second example:

$\text{[math]}$

Points $[1, 4, 5, 6, 7]$ are covered by one segment, points $[2, 3]$ are covered by two segments and there are no points covered by three segments.


#include<bits/stdc++.h>
 
using namespace std;
 
const int MAXN = 200000 + 10;
struct f
{
	long long head;
	int shu;
}a[2 * MAXN];
long long mark[MAXN];
 
int com(const f &u , const f &v)
{
	if(u.head == v.head)
		return u.shu > v.shu;
	return u.head < v.head;
}
 
int main()
{
	int n;
	cin >> n;
	int num = 0;
	for(int i = 1; i <= n * 2; i ++)
	{
		cin >> a[i].head;
		if(i & 1)
			a[i].shu = 1;
		else
			a[i].shu = 0;
	}
	//a[n + 1].head = a[n + 1].tail = 1e18 + 1;
	//cout << endl;
	sort( a + 1, a + 2 * n + 1, com);
	//for(int i = 1 ; i <= 2 * n ; i ++)
	//	cout << a[i].head << " " << a[i].shu << endl;
	//cout << endl;
	int  temp ;
	long long  f = 0;
	for(int i = 1 ; i <= n * 2; i ++)
	{
		//mark[num] += a[i].head - a[i - 1].head;
		if(a[i].shu == 1)
		{
			if(a[i].head - f - 1 >=0)
				mark[num] += a[i].head - f - 1;
			f = a[i].head;
			temp = i;
			int t = num;
			//cout << "!" << " " << t << endl;
			while(a[temp].head == a[temp + 1].head && temp + 1 <= n * 2)
			{
				//cout << a[temp + 1].shu << endl;
 
 
				//t ++;
				if(a[temp + 1].shu > 0)
				{
					num ++;
					t ++;
					//cout << "!" << endl;
				}
					//num ++;
				else
					num --;
				temp ++;
			}
			//cout << t << endl;
			num ++;
			t ++;
			//cout << num << " " << t << endl;
			mark[t] ++;
			i += (temp - i);
			//for(int j = 1; j <= 3 ; j ++)
			//	cout << mark[j] << " ";
			//cout << endl;
		}
		else
		{
			if(a[i].head - f - 1 >= 0)
				mark[num] += a[i].head - f - 1;
			f = a[i].head;
			int t = num ;
			temp = i;
			while(a[temp].head == a[temp + 1].head && temp + 1 <= n * 2)
			{
				if(a[temp + 1].shu > 0)
				{
					num ++;
					t ++;
				}
				//	num ++;
				else
					num --;
				temp ++;
			}
			//cout << num << " " << t << endl;
			mark[t] ++ ;
			num --;
 
			//for(int j = 1 ; j <= 3; j ++)
			//	cout << mark[j] << " ";
			//cout << endl;
			i += (temp - i) ;
		}
	}
	cout << mark[1];
	for(int i = 2; i <= n ; i ++)
		cout << " " << mark[i];
	return 0;
}

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