Codeforces1000C. Covered Points Count

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本文介绍了解决 CodeForces 平台上的 1000C 题目的一种方法。该题目要求计算在坐标轴上由多个区间覆盖的不同整数点的数量。通过记录每个区间的起始和结束位置，并对这些位置进行特殊处理和排序，最终实现了高效求解。

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http://codeforces.com/problemset/problem/1000/C

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn 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∈[1..n]k∈[1..n], calculate the number of points with integer coordinates such that the number of segments that cover these points equals kk. A segment with endpoints lili and riri covers point xx if and only if li≤x≤rili≤x≤ri.

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of segments.

The next nn lines contain segments. The ii-th line contains a pair of integers li,rili,ri (0≤li≤ri≤10180≤li≤ri≤1018) — the endpoints of the ii-th segment.

Output

Print nn space separated integers cnt1,cnt2,…,cntncnt1,cnt2,…,cntn, where cnticnti is equal to the number of points such that the number of segments that cover these points equals to ii.

Examples

input

Copy

output

Copy

6 2 1

input

Copy

output

Copy

5 2 0

Note

The picture describing the first example:

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

The picture describing the second example:

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

感觉思想很简单的一道题，就是实现的话...像我这样蒻的不好实现，把遇到的起点都赋值为1，终点都赋值为-1，然后记录每一个点，排序，有一个起点和终点重合的，会没法记录最大重合点，因为排序原因，先走的是负数，所以不肯能先加后面的起点，而是先减，所以就需要在终点后面+1，防止起点终点重合，还有就是数据过大，用ll，而且结构体要开二倍的。

#include<iostream>
#include<algorithm>
#include<map>
#include<string.h>
#define maxn 200050
#define ll long long
using namespace std;
map<int,int>last;
map<int,int>star;
struct node
{
	ll num;
	ll d;
}ac[maxn*2];
ll ans[maxn];
int cmp(node a,node b)
{
	if(a.num==b.num)
	return a.d<b.d;
	return a.num<b.num;
}
int main()
{
	ll n,total=0;
	cin>>n;
	ll x,y;
	memset(ans,0,sizeof(ans));
	for(ll i=1;i<=n;i++)
	{
		cin>>x>>y;
		ac[++total].num=x; ac[total].d=1;
		ac[++total].num=y+1; ac[total].d=-1;
	}
	sort(ac+1,ac+total+1,cmp);
	ll now=0;
	for(ll i=1;i<=total;i++)
	{
		ans[now]+=ac[i].num-ac[i-1].num;
		now+=ac[i].d;
	}
	for(ll i=1;i<=n;i++)
	cout<<ans[i]<<endl;
	return 0;
}