代码源 Div1 - 108#464. 数数（主席树，区间比k小的数的个数）HDU4417

Posted 2022-04-23 小哈里

problem

solution

• 主席树查询区间比k小的数的个数
• 建树之后直接在目标区间的主席树内将 H 作为挡板递归计数。

#include<bits/stdc++.h>
using namespace std;

#define ios ios::sync_with_stdio(0), cin.tie(0),cout.tie(0)
typedef long long LL;
const int maxn = 100010;

//离散化
int a[maxn], b[maxn], num;
int getidx(int v) return lower_bound(b+1,b+num+1,v)-b; 

//主席树,查询区间比k小的数的个数
struct HJTTree
    typedef long long LL;
    #define maxn 100010
    struct node int sum, l, r; hjt[maxn*40];
    int rt[maxn], cnt = 0;
    void init() cnt = 0; 
    inline int CreateNode(int sum, int l, int r)
        int idx = ++cnt;
        hjt[idx].sum = sum;
        hjt[idx].l = l;
        hjt[idx].r = r;
        return idx;
    
    void insert(int &p, int pp, int pos, int l, int r)
        p = CreateNode(hjt[pp].sum+1, hjt[pp].l, hjt[pp].r);
        if(l == r)return ;
        int mid = l+r>>1;
        if(pos <= mid)insert(hjt[p].l, hjt[pp].l, pos, l, mid);
        else insert(hjt[p].r, hjt[pp].r, pos, mid+1,r);
    
    int query(int u, int v, int h, int l, int r)
        if(r <= h)return hjt[v].sum-hjt[u].sum;
        int mid = l+r>>1;
        if(h <= mid)return query(hjt[u].l, hjt[v].l, h, l, mid);
        else return query(hjt[u].r,hjt[v].r, h, mid+1, r)+hjt[hjt[v].l].sum-hjt[hjt[u].l].sum;
    
hjt;

int main()
    IOS;
    int T;  cin>>T;
    while(T--)
        int n, m;  cin>>n>>m;
        for(int i = 1; i <= n; i++)
            cin>>a[i];  b[i] = a[i];
        
        sort(b+1,b+n+1);
        num = unique(b+1,b+n+1)-(b+1);
        hjt.init();
        for(int i = 1; i <= n; i++)
            hjt.insert(hjt.rt[i], hjt.rt[i-1], getidx(a[i]), 1, num);
        
        while(m--)
            int l, r, h;  cin>>l>>r>>h;
            // l++; r++;
            h = upper_bound(b+1,b+num+1,h)-b;
            if(h != 1)cout<<hjt.query(hjt.rt[l-1], hjt.rt[r], h-1, 1, num)<<"\\n";
            else cout<<"0\\n";
        
    
    return 0;