Code Forces 652D Nested Segments（离散化+树状数组）

 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

离散化+树状数组。

把所有区间按照右端点排序，然后统计左端点和右端点之间的已经包含的左端点个数，用树状数组求区间和会很快

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;
#define MAX 2*100000
struct Node
{
    int l,r;
    int pos;
}a[MAX+5];
int n;
int num[2*MAX+5];
int c[2*MAX+5];
int ans[MAX+5];
int s;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int num)
{
    while(x<=s)
    {
        c[x]+=num;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    int _sum=0;
    while(x>0)
    {
        _sum+=c[x];
        x-=lowbit(x);
    }
    return _sum;
}
int cmp(Node a,Node b)
{
    return a.r<b.r;
}
int main()
{
    scanf("%d",&n);
    memset(c,0,sizeof(c));
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].l,&a[i].r);
        a[i].pos=i;
        num[cnt++]=a[i].l;
        num[cnt++]=a[i].r;
    }
    sort(num,num+cnt);
    for(int i=1;i<=n;i++)
    {
        a[i].l=lower_bound(num,num+cnt,a[i].l)-num+1;
        a[i].r=lower_bound(num,num+cnt,a[i].r)-num+1;
    }
    sort(a+1,a+n+1,cmp);
    s=a[n].r;
    for(int i=1;i<=n;i++)
    {
        int num=sum(a[i].r)-sum(a[i].l-1);
        ans[a[i].pos]=num;
        update(a[i].l,1);
    }
    for(int i=1;i<=n;i++)
    {
        printf("%d
",ans[i]);
    }
    return 0;
}