Nested Segments CodeForces

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.

Example
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

/*    树状数组，离散化    对于原始段，按照左端点从大到小排序，如果左端点相同，则按照右端点从小到大进行排序，    这样的话，对于段 i 来说，可能包括它的段就是 x ( x > i ) 段；    要求段 i 包含了多少段，那么可以将每个段的右端点赋值 1，    于是排好序后，对于每个段，统计下它右端点前的和，即为结果，然后更新     但是，右端点的值很大，所以用按从小到大对应一个小的值，也就是离散化； */#include<bits/stdc++.h>using namespace std;const int maxn=200007*4;int sum[maxn];struct node{    int x,y,id;};node a[maxn];int ans[maxn];int n;int low_bit(int x){    return x&(-x);}int update(int x){    while(x<=n)    {        sum[x]++;        x+=low_bit(x);    }}int query(int x){    int ans=0;    while(x>0)    {        ans+=sum[x];        x-=low_bit(x);    }    return ans;}int cmp(node a,node b){    return a.y < b.y;}int cmp1(node a,node b){    if(a.x != b.x )        return a.x > b.x;    else return a.y < b.y;}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d%d",&a[i].x,&a[i].y);        a[i].id=i;    }    sort(a+1,a+1+n,cmp);    for(int i=1;i<=n;i++)        a[i].y=i;    sort(a+1,a+1+n,cmp1);    for(int i=1;i<=n;i++)    {        ans[a[i].id] = query(a[i].y);        update(a[i].y);    }    for(int i=1;i<=n;i++)        printf("%d\n",ans[i]);    return 0;}