CodeForces - 652D Nested Segments

Nested Segments

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 l_i and r_i ( - 10⁹ ≤ l_i < r_i ≤ 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 a_j — 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

题意：给出n条线段，求线段包含情况，输出包含数。
思路：使用树状数组，数据太大，需要离散化。然后按左端点排序即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn=400010;
int a[maxn],tree[maxn],n,ans[maxn];
struct node{
    int l,r,id;
    bool friend operator < (const node &u,const node &v){
        if(u.l==v.l) return u.r<v.r;
        return u.l>v.l;
    }
} p[maxn];
int lowbit(int k){
    return k&(-k);
}
void updata(int k){
    while(k<maxn){
        tree[k]+=1;
        k+=lowbit(k);
    }
}
int sum(int k){
    int ans=0;
    while(k){
        ans+=tree[k];
        k-=lowbit(k);
    }
    return ans;
}
int main(){
    int cnt=0;
    scanf("%d",&n);
    for(int i=1; i<=n; i++){
        scanf("%d %d",&p[i].l,&p[i].r);
        a[++cnt]=p[i].r;
        p[i].id=i;
    }
    sort(a+1,a+1+cnt);
    cnt=unique(a+1,a+cnt+1)-(a+1);
    for(int i=1; i<=n; i++) p[i].r=lower_bound(a+1,a+1+cnt,p[i].r)-a;
    sort(p+1,p+1+n);
    for(int i=1; i<=n; i++){
        ans[p[i].id]=sum(p[i].r+1);
        updata(p[i].r+1);
    }
    for(int i=1; i<=n; i++)
        printf("%d\n",ans[i]);
}

 

 

 

 

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~