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

41 82 34 75 6

output

3010

input

33 41 52 6

output

011

题意:给你n条线段(n<=2*10^5),每条线段一个[l,r],代表他的左右区间,输出对于每条线段,有多少条线段能被他完全覆盖。

 

思路:对r进行离散化,然后按L对查询从大到小进行排序,对每个r查询小于等于他的r有多少个

(树状数组求前缀和)

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef __int64 ll;const int maxn=200100;int C[maxn],Hash[maxn*2],ans[maxn];struct Point{    int x,y,id;}a[maxn];int n;bool cmp(Point u,Point v){    if(u.x!=v.x)        return u.x>v.x;    return u.y>v.y;}void add(int x){    while(x<=n){        C[x]+=1;        x+=(x&-x);    }}int sum(int x){    int ret=0;    while(x>0){        ret+=C[x];        x-=(x&-x);    }    return ret;}int main(){    int tot=0;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d%d",&a[i].x,&a[i].y);        a[i].id=i;        Hash[++tot]=a[i].y;    }    sort(Hash+1,Hash+tot+1);    int m=unique(Hash+1,Hash+tot+1)-Hash;    for(int i=1;i<=n;i++)        a[i].y=lower_bound(Hash+1,Hash+m,a[i].y)-Hash;    sort(a+1,a+n+1,cmp);    for(int i=1;i<=n;i++){        ans[a[i].id]=sum(a[i].y);        add(a[i].y);    }    for(int i=1;i<=n;i++)        printf("%d\n",ans[i]);    return 0;}