A - Jam's problem again HDU - 5618

Problem Description

Jam like to solve the problem which on the 3D-axis,given N(1≤N≤100000) points (x,y,z)(1≤x,y,z≤100000)

If two point such as$ (x_i,y_i,z_i)$ and \((x_j,y_j,z_j)\) \(x_i≥x_j,y_i≥_j,z_i≥z_j​\), the bigger one level add 1

Ask for the each level of the point.

Input

The first line is T(1≤T≤15) means T Case

For each case

The first line is N means the number of Point and next there are N line, each line has (x,y,z)

Output

Output with N line,each line has one number means the lever of point

Sample Input

1
4
10 4 7
10 6 6
8 2 5
7 3 10

Sample Output

1
1
0
0

题解

三维偏序

初始按\(x\)为意义关键字,\(y\)为第二关键字,\(z\)为第三关键字排序,一定不能省只按\(x\)排序,因为要处理相同的点

分别处理子问题,然后以\(y\)坐标为关键字, 将左右两端归并排序(这样递归左右内部\(y\)有序)

并用树状数组维护前缀的\(z\)的信息(当前左边的\(x\)一定比右边小,并且\(y\)也有序)

归并到左边的点就插入\(z\), 右边的点询问\(z\)

对于相同的点,我们初始发现排完序后,相同的点中序号较前的点不会统计到后面相同的点,只有最后的那个点才统计完全

所以可以在一开始先将相同的点加上\(p, p-1, ..., 1, 0\)就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>using namespace std;typedef long long LL;
const int MAXN = 1e5 + 10;inline LL in()
{LL x = 0, flag = 1; char ch = getchar();while (ch < '0' || ch > '9') { if (ch == '-') flag = -1; ch = getchar(); }while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();return x * flag;
}int T;int n;struct Fenwick
{int val[MAXN];void clear() { memset(val, 0, sizeof val); }void update(int x, int v){for (int i = x; i <= 100000; i += i & (-i)) val[i] += v;}int query(int x){int ret = 0;for (int i = x; i > 0; i &= (i - 1)) ret += val[i];return ret;}
} FT;int an[MAXN];
struct Node
{int x, y, z, id;bool operator < (const Node & b) const{return x == b.x ? (y == b.y ? z < b.z : y < b.y) : x < b.x; }
} a[MAXN], b[MAXN];void merge(int l, int mid, int r)
{int i = l, j = mid + 1, k = i;while (k <= r){if (j > r || (i <= mid && a[i].y <= a[j].y)){FT.update(a[i].z, 1);b[k ++] = a[i];++ i;}else{an[a[j].id] += FT.query(a[j].z);b[k ++] = a[j];++ j;}}for (int i = l; i <= mid; i ++) FT.update(a[i].z, -1);for (int i = l; i <= r; i ++) a[i] = b[i];
}
void solve(int l, int r)
{if (l >= r) return;int mid = (l + r) >> 1;solve(l, mid);solve(mid + 1, r);merge(l, mid, r);
}int main()
{T = in();while (T --){memset(a, 0, sizeof a);memset(an, 0, sizeof an);n = in(); for (int i = 1; i <= n; i ++) a[i] = (Node) { in(), in(), in(), i };sort(a + 1, a + n + 1);for (int i = n - 1; i >= 1; i --){if (a[i].x == a[i + 1].x && a[i].y == a[i + 1].y && a[i].z == a[i + 1].z) an[a[i].id] += an[a[i + 1].id] + 1;}solve(1, n);for (int i = 1; i <= n; i ++) printf("%d\n", an[i]);}return 0;
}/**/