[Educational Codeforces Round 10D] [트 리 배열] Nested Segments 각 라인 내부 에 몇 개의 라인 이 있 습 니까?

D. Nested Segments
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.
Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.
Each of the next n lines contains two integers li and ri ( - 109 ≤ li < ri ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.
Output
Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.
Examples
input

4
1 8
2 3
4 7
5 6

output

3
0
1
0

input

3
3 4
1 5
2 6

output

0
1
1

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 2e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
int b[N + N];
int l[N], r[N];
map<int, int>mop;
map<int, int>::iterator it;
struct A
{
	int x, y, o;
	bool operator < (const A& b)const
	{
		if (x != b.x)return x < b.x;
		return y < b.y;
	}
}a[N];

int id;
void add(int x, int val)
{
	for (x; x <= id; x += (x&-x))
	{
		b[x] += val;
	}
}
int check(int x)
{
	int tmp = 0;
	for (; x; x -= (x&-x))
	{
		tmp += b[x];
	}
	return tmp;
}

int ans[N];
int main()
{
	while (~scanf("%d", &n))
	{
		mop.clear();
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d%d", &l[i], &r[i]);
			mop[l[i]] = 0;
			mop[r[i]] = 0;
		}

		id = 0;
		for (it = mop.begin(); it != mop.end(); ++it)
		{
			it->second = ++id;
			b[id] = 0;
		}

		for (int i = 1; i <= n; ++i)
		{
			int x = mop[l[i]];
			int y = mop[r[i]];
			a[i].x = x;
			a[i].y = y;
			a[i].o = i;
			add(y, 1);
		}
		sort(a + 1, a + n + 1);

		int p = 1;
		for (int i = 1; i <= n; ++i)
		{
			while (p <= n&&a[p].x < a[i].x)
			{
				add(a[p].y, -1);
				++p;
			}
			ans[a[p].o] = check(a[p].y) - 1;
		}
		for (int i = 1; i <= n; ++i)printf("%d
", ans[i]);
	}
	return 0;
}
/*
【  】
 n（2e5）     ，     [-1e9,1e9]
            （         ）
              

【  】
    

【  】
              。
             。            <=          。

【     &&  】
O(nlogn)

*/