題意：詢問區間[l,r]內有幾個數字小於h

思路：對於每次詢問H,L,R，僅需要考慮比H小的hi在L,R範圍內的個數；

   為了避免比H大的hi的影響，可以想到對詢問H，和hi進行排序

   在詢問H時，僅將比H小的hi插入到樹狀陣列中

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 100002
#define lowbit(i) (i & (-i))
int prefix[MAXN],ans[MAXN];
struct Interval
{
    int h,index;
    bool operator < (const Interval & inter)const{
        return h < inter.h;
    }
}interval[MAXN];
struct Query
{
    int l,r,h,id;
    bool operator < (const Query & qu)const{
        return h < qu.h;
    }
}query[MAXN];
int sum(int i)
{
    int total = 0;
    while(i)
    {
        total += prefix[i];
        i -= lowbit(i);
    }
    return total;
}
void modify(int i,int maxn,int key)
{
    while(i <= maxn)
    {
        prefix[i] += key;
        i += lowbit(i);
    }
}
int main()
{
    int _,n,m,ql,qr,val;
    scanf("%d",&_);
    for(int kcase = 1;kcase <= _;kcase++)
    {
        printf("Case %d:\n",kcase);
        scanf("%d%d",&n,&m);
        memset(prefix,0,sizeof(prefix));
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&interval[i].h);
            interval[i].index = i;
        }
        for(int i = 1;i <= m;i++)
        {
            scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].h);
            query[i].id = i;
        }
        sort(interval + 1,interval + n + 1);
        sort(query + 1,query + m + 1);
        for(int j = 1,i = 1;i <= m;i++)
        {
            while(j != m + 1 && interval[j].h <= query[i].h)
            {
                modify(interval[j].index,n,1);
                j++;
            }
            ans[query[i].id] = sum(query[i].r + 1) - sum(query[i].l);
        }
        for(int i = 1;i <= m;i++)
            printf("%d\n",ans[i]);
    }
}
 

也可以用二分加劃分樹來做

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 100001
#define LOG 20
int sorted[MAXN] , part_tree[MAXN][LOG] , left_cnt[MAXN][LOG];
int ql,qr;
void build(int l,int r,int deep)
{
    if(l == r)return;
    int mid = l + r >> 1 , same = left_cnt[0][deep] = 0;
    for(int i = mid;i >= l;i--)
        if(sorted[mid] == sorted[i])same++;
        else break;
    int lcnt = l , rcnt = mid + 1;
    for(int i = l;i <= r;i++)
    {
        if(part_tree[i][deep] < sorted[mid])
            part_tree[lcnt++][deep + 1] = part_tree[i][deep];
        else if(part_tree[i][deep] == sorted[mid] && same)
        {
            part_tree[lcnt++][deep + 1] = part_tree[i][deep];
            same--;
        }
        else
            part_tree[rcnt++][deep + 1] = part_tree[i][deep];
        left_cnt[i][deep] = lcnt - l + left_cnt[l - 1][deep];
    }
    build(l,mid,deep + 1) , build(mid + 1,r,deep + 1);
}
int query(int l,int r,int deep,int ql,int qr,int kth)
{
    if(ql == qr)return part_tree[ql][deep];
    int mid = l + r >> 1;
    int lcnt = left_cnt[qr][deep] - left_cnt[ql - 1][deep];
    if(lcnt >= kth)
    {
        ql = l + left_cnt[ql - 1][deep] - left_cnt[l - 1][deep];
        qr = ql + lcnt - 1;
        return query(l,mid,deep + 1,ql,qr,kth);
    }
    else
    {
        qr += left_cnt[r][deep] - left_cnt[qr][deep];
        ql += left_cnt[r][deep] - left_cnt[ql - 1][deep];
        return query(mid + 1,r,deep + 1,ql,qr,kth - lcnt);
    }
}
int binary(int val,int n)
{
    int l = 1 , r = qr - ql + 1;
    if(query(1,n,0,ql + 1,qr + 1,1) > val)return 0;
    while(l < r)
    {
        int mid = l + r + 1 >> 1;
        if(query(1,n,0,ql +1,qr + 1,mid) <= val)l = mid;
        else r = mid - 1;
    }
    return r;
}
int main()
{
    int _,n,m,kth,val;
    scanf("%d",&_);
    for(int kcase = 1;kcase <= _;kcase++)
    {
        printf("Case %d:\n",kcase);
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&part_tree[i][0]);
            sorted[i] = part_tree[i][0];
        }
        sort(sorted + 1,sorted + n + 1);
        build(1,n,0);
        while(m--)
        {
            scanf("%d%d%d",&ql,&qr,&val);
            int ans = binary(val,n);
            printf("%d\n",ans);
        }
    }
}
 

線段樹每個節點維護一段有序序列

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
#define MAXN 100002
const int root = 1;
vector<int>vec[MAXN << 2];
int h[MAXN],ql,qr;
void build(int l,int r,int cnt)
{
    vec[cnt].clear();
    for(int i = l;i <= r;i++)
        vec[cnt].push_back(h[i]);
    sort(vec[cnt].begin(),vec[cnt].end());
    int mid = l + r >> 1;
    if(l == r)return;
    build(l,mid,cnt << 1);
    build(mid + 1,r,cnt << 1 | 1);
}
int query(int l,int r,int cnt,int key)
{
    if(ql <= l && r <= qr)
        return upper_bound(vec[cnt].begin(),vec[cnt].end(),key) - vec[cnt].begin();
    int mid = l + r >> 1 , ret = 0;
    if(ql <= mid)ret = query(l,mid,cnt << 1,key);
    if(qr >= mid + 1)ret += query(mid + 1,r,cnt << 1 | 1,key);
    return ret;
}
int main()
{
    int _,n,m,val;
    scanf("%d",&_);
    for(int kcase = 1;kcase <= _;kcase++)
    {
        printf("Case %d:\n",kcase);
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++)
            scanf("%d",&h[i]);
        build(1,n,root);
        for(int i = 1;i <= m;i++)
        {
            scanf("%d%d%d",&ql,&qr,&val);
            ql++ , qr++;
            int ans = query(1,n,root,val);
            printf("%d\n",ans);
        }
    }
}