Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input 1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3
Sample Output Case 1: 4 0 0 3 1 2 0 1 5 1

將磚塊高度val和瑪麗能跳的高度H都存起來，更新p[k].val<=q[i].h的，區間求和的時候就能保證當前更新的都是小於瑪麗能跳的的高度。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
const int maxn=1e5+10;
struct tree{
    int l,r;
    int cnt;
}t[maxn<<2];
struct node1{
    int val;
    int pos;
    bool operator<(const node1 l1)const{
       return val<l1.val;
    }
}p[maxn];//儲存瑪麗的原始位置和能跳的高度
struct node2{
    int l,r;
    int id,h;
    bool operator<(const node2 l2)const{
        return h<l2.h;
    }
}q[maxn];//儲存m次詢問
int ans[maxn];
void pushup(int rs)
{
    t[rs].cnt=t[rs<<1].cnt+t[rs<<1|1].cnt;
}
void build(int rs,int l,int r)
{
    t[rs].l=l;
    t[rs].r=r;
    t[rs].cnt=0;
    if(l==r)
        return ;
    int mid=(l+r)>>1;
    build(rs<<1,l,mid);
    build(rs<<1|1,mid+1,r);
}
void update(int rs,int pos)
{
     t[rs].cnt++;
     if(t[rs].l==t[rs].r)
        return ;
    int mid=(t[rs].l+t[rs].r)>>1;
    if(pos<=mid)  update(rs<<1,pos);
    else   update(rs<<1|1,pos);
}
int query(int l,int r,int rs)
{
    if(t[rs].l==l&&t[rs].r==r)
        return t[rs].cnt;
    int mid=(t[rs].l+t[rs].r)>>1;
    if(r<=mid)   return query(l,r,rs<<1);
    else if(l>mid)  return query(l,r,rs<<1|1);
    else  return query(l,mid,rs<<1)+query(mid+1,r,rs<<1|1);
}
int main()
{
    int t,n,m;
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        build(1,1,n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&p[i].val);
            p[i].pos=i;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h);
            q[i].id=i;
        }
        sort(p+1,p+1+n);
        sort(q+1,q+1+m);
        int k=1;
        for(int i=1;i<=m;i++)
        {
            while(k<=n&&p[k].val<=q[i].h)
            {
                update(1,p[k].pos);
                k++;
            }
            ans[q[i].id]=query(q[i].l+1,q[i].r+1,1);
        }
        printf("Case %d:\n",cas++);
        for(int i=1;i<=m;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}
 

另外樹狀陣列也可以做。類似的單點更新和區間求和問題都可以用樹狀陣列來寫。

類似離線線段樹的思想將節點和詢問儲存排序後依次更新，就能保證結果的正確性。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
const int maxn=1e5+10;
int cnt[maxn];
struct node1{
    int val;
    int pos;
    bool operator<(const node1 l1)const{
       return val<l1.val;
    }
}p[maxn];//儲存瑪麗的原始位置和能跳的高度
struct node2{
    int l,r;
    int id,h;
    bool operator<(const node2 l2)const{
        return h<l2.h;
    }
}q[maxn];//儲存m次詢問
int ans[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x)
{
    while(x<maxn)
    {
        cnt[x]++;
        x+=lowbit(x);
    }
}
int query(int x)
{
    int s=0;
    while(x>0)
    {
        s+=cnt[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int t,n,m;
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        memset(cnt,0,sizeof(cnt));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&p[i].val);
            p[i].pos=i;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h);
            q[i].id=i;
        }
        sort(p+1,p+1+n);
        sort(q+1,q+1+m);
        int k=1;
        for(int i=1;i<=m;i++)
        {
            while(k<=n&&p[k].val<=q[i].h)
            {
                update(p[k].pos);
                k++;
            }
            ans[q[i].id]=query(q[i].r+1)-query(q[i].l);
        }
        printf("Case %d:\n",cas++);
        for(int i=1;i<=m;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}