L - A Heap of Heaps CodeForces - 538F 主席树

L - A Heap of Heaps

 CodeForces - 538F 

这个是一个还比较裸的静态主席树。

这个题目的意思是把这个数组变成k叉树，然后问构成的树的子树小于等于它的父节点的对数有多少。

因为这个k是从1~n-1 所以直接暴力肯定是不对的，所以可以用主席树来查询区间第k大。

查询的次数大约为n+n/2+n/3+...n/n 差不多是n*log(n) 的复杂度，建个主席树，直接查询即可

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 2e5 + 10;
int n, m, root[maxn], a[maxn], b[maxn], cnt;
int sum[maxn << 5], lc[maxn << 5], rc[maxn << 5];

void build(int &rt, int l, int r) {
    rt = ++cnt;
    sum[rt] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(lc[rt], l, mid);
    build(rc[rt], mid + 1, r);
    // printf("rt=%d l=%d r=%d\n",rt,l,r);
}

int update(int rt, int l, int r, int pos) {
    // printf("ww  rt=%d l=%d r=%d pos=%d\n", rt, l, r, pos);
    int id = ++cnt;
    sum[id] = sum[rt] + 1;
    // printf("rt=%d sum[%d]=%d\n", rt, id, sum[id]);
    lc[id] = lc[rt], rc[id] = rc[rt];
    if (l == r) return id;
    int mid = (l + r) >> 1;
    // printf("mid=%d rt=%d l=%d r=%d pos=%d\n", mid,rt,l,r,pos);
    if (pos <= mid) lc[id] = update(lc[id], l, mid, pos);
    else rc[id] = update(rc[id], mid + 1, r, pos);
    // printf("rt=%d l=%d r=%d pos=%d\n", rt, l, r, pos);
    return id;
}

int query(int l, int r, int u, int v, int h) {
    int mid = (l + r) >> 1;
    int x = sum[lc[v]] - sum[lc[u]];
    //printf("l=%d r=%d u=%d v=%d h=%d mid=%d x=%d  %d\n", l, r, u, v, h, mid, x, sum[v] - sum[u]);
    if (l == r) return sum[v] - sum[u];
    //printf("ww l=%d r=%d u=%d v=%d h=%d mid=%d x=%d\n", l, r, u, v, h, mid, x);
    int ans = 0;
    if (h <= mid) ans = query(l, mid, lc[u], lc[v], h);
    else ans = x + query(mid + 1, r, rc[u], rc[v], h);
    return ans;
}

int main() {
    cnt = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + 1 + n);
    int len = unique(b + 1, b + 1 + n) - b - 1;
    //printf("len=%d\n", len);
    build(root[0], 1, len);
    for (int i = 1; i <= n; i++) {
        a[i] = lower_bound(b + 1, b + 1 + len, a[i]) - b;
    //    printf("a[%d]=%d\n", i, a[i]);
        root[i] = update(root[i - 1], 1, len, a[i]);
    }
    // if(len==1)
    // {
    //     for (int i = 1; i < n; i++) printf("0 ");
    //     printf("\n");
    //     return 0;
    // }
    for (int i = 1; i <= n - 1; i++) {
        int j = 1, ans = 0;
        while (i*j + 1 <= n) {
            int l = i * (j - 1) + 1, r = i * j + 1;
            if (a[j] - 1 != 0) ans += query(1, len, root[l], root[r], a[j] - 1);
            j++;
        }
        if (i*j + 1 > n&&i*(j - 1) + 2 <= n) {
            int l = i * (j - 1) + 1, r = n;
            if (a[j] - 1 != 0) ans += query(1, len, root[l], root[r], a[j] - 1);
        }
        printf("%d ", ans);
    }
    printf("\n");
    return 0;
}

标签：Heaps,rt,int,sum,CodeForces,mid,538F,printf,id
来源： https://www.cnblogs.com/EchoZQN/p/11391462.html