Educational Codeforces Round 10 D. Nested Segments

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

离散化+ 线段树或树状数组维护前缀和

想按照左端点进行排序，然后逆序 对右端点进行查询,插入.这样可以保证线段的包含关系。写的有点挫，一开始始终re后来把线段树开到8*maxn就过了。。

/* ***********************************************
Author        :guanjun
Created Time  :2016/3/28 15:34:10
File Name     :cfedu10d.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 211000
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
struct Node{
    int x,y;
    int id;
}Nod[maxn];
bool cmp(Node a,Node b){
    return a.x<b.x;
}
struct node{
    int l,r;
    ll sum;
    int c;
}nod[maxn*];
int ans[maxn];
void push_up(int i){
    nod[i].sum=nod[i<<].sum+nod[i<<|].sum;
}
void build(int i,int l,int r){
    nod[i].l=l;
    nod[i].r=r;
    nod[i].c=;
    if(l==r){
        nod[i].sum=;
        return ;
    }
    int mid=(l+r)/;
    build(i<<,l,mid);
    build(i<<|,mid+,r);
    push_up(i);
}
void update(int i,int k,int x){
    if(nod[i].l==k&&nod[i].r==k){
        nod[i].sum+=x;
        return ;
    }
    int mid=(nod[i].l+nod[i].r)/;
    if(k<=mid)update(i<<,k,x);
    else update(i<<|,k,x);
    push_up(i);
}
ll quary(int i,int l,int r){
    if(nod[i].l==l&&nod[i].r==r){
        return nod[i].sum;
    }
    int mid=(nod[i].l+nod[i].r)/;
    ll ans=;
    if(r<=mid)ans+=quary(i<<,l,r);
    else if(l>mid)ans+=quary(i<<|,l,r);
    else {
        ans+=quary(i<<,l,mid);
        ans+=quary(i<<|,mid+,r);
    }
    push_up(i);
    return ans;
}
int num[*maxn];
map<int,int>mp;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,q;
    ll z;
    while(cin>>n){
        mp.clear();
        for(int i=;i<=n;i++){
            scanf("%d%d",&Nod[i].x,&Nod[i].y);
            num[i]=Nod[i].x;
            num[i+n]=Nod[i].y;
            Nod[i].id=i;
        }
        sort(num+,num++(*n));
        int cnt=;
        for(int i=;i<=*n;i++){
            mp[num[i]]=cnt;
            cnt++;
        }
        build(,,mp[num[*n]]);
        sort(Nod+,Nod++n,cmp);
        for(int i=n;i>=;i--){
            ans[Nod[i].id]=quary(,,mp[Nod[i].y]);
            update(,mp[Nod[i].y],);
        }
        for(int i=;i<=n;i++){
            printf("%d\n",ans[i]);
        }
    }
    return ;
}