題目鏈接

思路

對一維排序後，使用$cdq$分治，以類似歸並排序的方法處理的二維，對於滿足$a[i].b \leq a[j].b$的點對，用樹狀數組維護$a[i].c$的數量。當遇到$a[i].b>a[j].b$時可以更新$j$的答案，因為前半部分中剩余的點的第二維必然大於$j$點的第二維（記住我們是對第二維進行歸並排序所以第二維是有序的，因此有這樣的判斷）。每次要記得初始化樹狀數組。

代碼

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl

using namespace std;
typedef long long LL;

const int N = 1e5 + 5;

int n, k;
LL ans[N];

struct node {
    int x, y, z, w, f;
    bool operator < (const node &a) const {
        if(x != a.x) return x < a.x;
        if(y != a.y) return y < a.y;
        return z < a.z;
    }
} a[N], b[N];

struct BIT {
    static const int M = 2e5 + 5;
    int c[M];

    int lowbit(int x) {return (x & (-(x)));}

    void update(int x, int y) {
        while(x < M) {
            c[x] += y;
            x += lowbit(x);
        }
    }

    int query(int x) {
        int res = 0;
        while(x) {
            res += c[x];
            x -= lowbit(x);
        }
        return res;
    }
} tree;

void solve(int L, int R) {
    if(L == R) return;
    int Mid = (L + R) / 2;
    solve(L, Mid); solve(Mid + 1, R);
    int t1 = L, t2 = Mid + 1, tot = L;
    for(int i = L; i <= R; i++) {
        if((t1 <= Mid && a[t1].y <= a[t2].y) || t2 > R) tree.update(a[t1].z, a[t1].w), b[tot++] = a[t1++];
        else a[t2].f += tree.query(a[t2].z), b[tot++] = a[t2++];
    }
    for(int i = L; i <= Mid; i++) tree.update(a[i].z, -a[i].w);
    for(int i = L; i <= R; i++) a[i] = b[i];
}

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z), a[i].w = 1;
    sort(a + 1, a + 1 + n);
    int cnt = 1;
    for(int i = 2; i <= n; i++) {
        if(a[i].x == a[cnt].x && a[i].y == a[cnt].y && a[i].z == a[cnt].z) a[cnt].w++;
        else a[++cnt] = a[i];
    }
    solve(1, cnt);
    for(int i = 1; i <= cnt; i++) ans[a[i].w + a[i].f - 1] += 1LL * a[i].w;
    for(int i = 0; i < n; i++) printf("%lld\n", ans[i]);
    return 0;
}

luogu P3810 三維偏序（陌上花開）CDQ分治