hdu4417 主席树求区间小于等于K

超、凢脫俗 2021-11-02 09:48 323阅读 0赞

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4417

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

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:

主席树板子题，求区间小于等于h的数有多少，注意题目给的查询区间是从0开始的

#include<iostream>
    #include<algorithm>
    using namespace std;
    #define ll long long
    #define maxn 100005
    int a[maxn],b[maxn],T[maxn<<5],L[maxn<<5],R[maxn<<5],sum[maxn<<5],tot;
    inline int update(int pre,int l,int r,int x)
    {
        int rt=++tot;
        L[rt]=L[pre];
        R[rt]=R[pre];
        sum[rt]=sum[pre]+1;
        if(l<r)
        {
            int mid=l+r>>1;
            if(x<=mid)L[rt]=update(L[pre],l,mid,x);
            else R[rt]=update(R[pre],mid+1,r,x);
        }
        return rt;
    }
    inline int query(int u,int v,int ql,int qr,int l,int r)
    {
        if(ql<=l&&qr>=r)return sum[v]-sum[u];
        int mid=l+r>>1,ans=0;
        if(ql<=mid)ans+=query(L[u],L[v],ql,qr,l,mid);
        if(qr>mid)ans+=query(R[u],R[v],ql,qr,mid+1,r);
        return ans;
    }
    int main()
    {
        int t;
        cin>>t;
        for(int W=1;W<=t;W++)
        {
            int n,m;
            cin>>n>>m;
            for(int i=1;i<=n;i++)
            {
                cin>>a[i];
                b[i]=a[i];
            }
            sort(b+1,b+1+n);
            int len=unique(b+1,b+1+n)-b-1;
            T[0]=sum[0]=L[0]=R[0]=tot=0;
            for(int i=1;i<=n;i++)
            {
                int pos=lower_bound(b+1,b+1+len,a[i])-b;
                T[i]=update(T[i-1],1,len,pos);
            }
            cout<<"Case "<<W<<":"<<endl;
            for(int i=1;i<=m;i++)
            {
                int l,r,h;
                cin>>l>>r>>h;
                int pos=upper_bound(b+1,b+1+len,h)-b;
                pos--;
                if(!pos)cout<<"0"<<endl;
                else
                {
                    cout<<query(T[l],T[r+1],1,pos,1,len)<<endl;
                }
            }
        }
        return 0;
    }

转载于:https://www.cnblogs.com/chen99/p/11293955.html