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 << 61;
const double eps=1e-5;
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*8];
int ans[maxn];
void push_up(int i){
nod[i].sum=nod[i<<1].sum+nod[i<<1|1].sum;
}
void build(int i,int l,int r){
nod[i].l=l;
nod[i].r=r;
nod[i].c=0;
if(l==r){
nod[i].sum=0;
return ;
}
int mid=(l+r)/2;
build(i<<1,l,mid);
build(i<<1|1,mid+1,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)/2;
if(k<=mid)update(i<<1,k,x);
else update(i<<1|1,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)/2;
ll ans=0;
if(r<=mid)ans+=quary(i<<1,l,r);
else if(l>mid)ans+=quary(i<<1|1,l,r);
else {
ans+=quary(i<<1,l,mid);
ans+=quary(i<<1|1,mid+1,r);
}
push_up(i);
return ans;
}
int num[2*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=1;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+1,num+1+(2*n));
int cnt=1;
for(int i=1;i<=2*n;i++){
mp[num[i]]=cnt;
cnt++;
}
build(1,1,mp[num[2*n]]);
sort(Nod+1,Nod+1+n,cmp);
for(int i=n;i>=1;i--){
ans[Nod[i].id]=quary(1,1,mp[Nod[i].y]);
update(1,mp[Nod[i].y],1);
}
for(int i=1;i<=n;i++){
printf("%d\n",ans[i]);
}
}
return 0;
}