luogu P3810 三维偏序(陌上花开)cdq分治
思路
对一维排序后,使用$cdq$分治,以类似归并排序的方法处理的二维,对于满足$a[i].b \leq a[j].b$的点对,用树状数组维护$a[i].c$的数量。当遇到$a[i].b>a[j].b$时可以更新$j$的答案,因为前半部分中剩余的点的第二维必然大于$j$点的第二维(记住我们是对第二维进行归并排序所以第二维是有序的,因此有这样的判断)。每次要记得初始化树状数组。
代码
#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; }