codeforcces 1000C 模拟+思维

C. Covered Points Count

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $k\in \left[\mathrm{1..}n\right]$, 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 $l_i\le x\le r_i$.

Input

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

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

Output

Print $n$ space separated integers $cn{t}_1,cn{t}_2,\dots ,cn{t}_n$, where $cn{t}_i$ 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:

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

The picture describing the second example:

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.

题意：有n个线段，每个线段头为ai,尾为bi，线段覆盖[ai,bi]的所有点。问被1~n条线段覆盖的整数点各有多少个。

思路：记录下每个是线段起点或终点的整数点的线段覆盖数的变化，注意，由于bi页也被其对应线段覆盖，所以在计算线段终点时，应取bi+1.

在第一例中，可得如下记录：0：+1；1：+1，3：+1；4：-2；9：-2。

该记录可以用map存储。

然后按顺序访问每个可能的起点和终点即可。

代码：

#include 

using namespace std;

int n;
typedef long long ll;
set  S;
map  M;
ll ans[400005];

int main()
{
    ll l,r;
    scanf("%d",&n);
    for (int i=0; i::iterator it;
    it = S.begin();
    ll pre = *it;
    it++;
    int cnt = M[pre];
    //cout<