#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<queue>#include<iostream>using namespace std;typedef long long LL;const int maxn = 2e5 + 10;const int low(int x){ return x&-x; }int T, n;struct point{    int x, y, z, id, cnt;    void read(){ scanf("%d%d%d", &x, &y, &z); cnt = 0; }    bool operator<(const point& a) const    {        if (x == a.x&&y == a.y) return z < a.z;        if (x == a.x) return y < a.y;        return x < a.x;    }}a[maxn];bool cmp(const point&a, const point&b){    return a.id < b.id;}struct Tree{    //将y坐标用树状数组维护，z坐标用线段树维护。然后在树状数组中的每一个点都对应着一个线段树。    //查询小于(y,z)的所有点只需要在1-y这个树状数组中的所有点对应的线段树中都查询<z的个数    int tot, first[maxn], f[maxn * 50], L[maxn * 50], R[maxn * 50], lit, lim;    //lim代表最大的z，lit代表最大的y    void clear(int x, int y)    {        lit = x;    lim = y;        for (int i = 1; i <= lit; i++)        {            first[i] = i;            f[i] = R[i] = L[i] = 0;        }        tot = lit + 1;    }    int newnode()    {        f[tot] = R[tot] = L[tot] = 0;        return tot++;    }    void ins(int x, int l, int r, int v)    {        f[x] += 1;        if (l == r) return;        int mid = l + r >> 1;        if (v <= mid) { if (!L[x]) L[x] = newnode(); ins(L[x], l, mid, v); }        else { if (!R[x]) R[x] = newnode(); ins(R[x], mid + 1, r, v); }    }    void insert(int x, int y)    {        for (int i = x; i <= lit; i += low(i)) ins(first[i], 1, lim, y);    }    int find(int x, int l, int r, int v)//对于第x棵线段树，查询有多少权值<v的    {        if (r <= v) return f[x];        int mid = l + r >> 1, ans = 0;        if (v <= mid) ans = find(L[x], l, mid, v);        else ans = f[L[x]] + find(R[x], mid + 1, r, v);        return ans;    }    int get(int x, int y)//通过树状数组统计1-x这个范围内的线段树中<y的点的总数    {        int res = 0;        for (int i = x; i; i -= low(i)) res += find(first[i], 1, lim, y);        return res;    }}tree;int main(){    scanf("%d", &T);    while (scanf("%d", &n) != EOF, T--)    {        for (int i = 0; i < n; i++) a[i].read(), a[i].id = i;        sort(a, a + n);//将输入的点按照题意排序//        for(int i=0;i<n;i++){//            cout<<a[i].x<<endl;//        }        for (int i = n - 2; i >= 0; i--)        if (a[i].x == a[i + 1].x&&a[i].y == a[i + 1].y&&a[i].z == a[i + 1].z) a[i].cnt = a[i + 1].cnt + 1;        int ans1 = 0, ans2 = 0;        for (int i = 0; i < n; i++) ans1 = max(ans1, a[i].y), ans2 = max(ans2, a[i].z);        tree.clear(ans1, ans2);        for (int i = 0; i < n; i++)//将三维转为2维，以为已经排好序，所以先插入的点的x坐标<后插入点的x坐标，接下来只需要比较y和z        {            a[i].cnt += tree.get(a[i].y, a[i].z);            tree.insert(a[i].y, a[i].z);        }        sort(a, a + n, cmp);        for (int i = 0; i < n; i++) printf("%d\n", a[i].cnt);    }    return 0;}

#include <cstdio>#include <cstring>#include <algorithm>#define Rep(i,n) for(int i = 1;i <= n;i ++)#define u t[x]using namespace std;const int N = 100005;int n,ans[N],t[N];struct Point {int x,y,z,id,sum;}p[N];bool cx(Point a,Point b){return a.x == b.x ? (a.y == b.y ? a.z < b.z : a.y < b.y) : a.x < b.x;}bool cy(Point a,Point b){    if(a.y != b.y)return a.y < b.y;    return a.x == b.x ? a.z < b.z : a.x < b.x;}void add(int x,int val){for(;x <= 100000;x += x & -x)u += val;}int Qry(int x){int s = 0;for(;x;x -= x & -x)s += u;return s;}void solve(int l,int r){    if(l == r)return;    int mid = l + r >> 1;    solve(l,mid),solve(mid + 1,r);    sort(p + l,p + mid + 1,cy);    sort(p + mid + 1,p + r + 1,cy);    int j = l;    for(int i = mid + 1;i <= r;i ++)    {        for(;j <= mid && p[j].y <= p[i].y;j ++)            add(p[j].z,1);        p[i].sum += Qry(p[i].z);    }    for(j --;j >= l;j --)add(p[j].z,-1);}int main (){    int _;scanf("%d",&_);    while(_--)    {        scanf("%d",&n);        Rep(i,n)scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z),p[i].id = i,p[i].sum = 0;        sort(p + 1,p + 1 + n,cx);        solve(1,n);        for(int i = 1;i <= n;)        {            int j = i + 1;            int tmp = p[i].sum;            for(;j <= n  && p[i].x == p[j].x && p[i].y == p[j].y && p[i].z == p[j].z ;++ j)tmp = max(tmp,p[j].sum);            for(int k = i;k < j;k ++)ans[p[k].id] = tmp;            i = j;        }        Rep(i,n)            printf("%d\n",ans[i]);    }    return 0;}