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

41 82 34 75 6

output

3010

input

33 41 52 6

output

011

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*100000struct 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\n",ans[i]);    }    return 0;}