• HDOJ 2665 Kth number



    静态区间第K小....划分树裸题

    Kth number

    Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 5341    Accepted Submission(s): 1733


    Problem Description
    Give you a sequence and ask you the kth big number of a inteval.
     

    Input
    The first line is the number of the test cases. 
    For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere. 
    The second line contains n integers, describe the sequence. 
    Each of following m lines contains three integers s, t, k. 
    [s, t] indicates the interval and k indicates the kth big number in interval [s, t]
     

    Output
    For each test case, output m lines. Each line contains the kth big number.
     

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

    Sample Output
    2
     

    Source
     


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    const int maxn=100100;
    
    int tree[20][maxn];
    int sorted[maxn];
    int toleft[20][maxn];
    
    void build(int l,int r,int dep)
    {
        if(l==r) return ;
        int mid=(l+r)/2;
        int same=mid-l+1;
        for(int i=l;i<=r;i++)
            if(tree[dep][i]<sorted[mid]) same--;
        int lpos=l,rpos=mid+1;
        for(int i=l;i<=r;i++)
        {
            if(tree[dep][i]<sorted[mid])
                tree[dep+1][lpos++]=tree[dep][i];
            else if(tree[dep][i]==sorted[mid]&&same>0)
            {
                tree[dep+1][lpos++]=tree[dep][i];
                same--;
            }
            else tree[dep+1][rpos++]=tree[dep][i];
    
            toleft[dep][i]=toleft[dep][l-1]+lpos-l;
        }
        build(l,mid,dep+1);
        build(mid+1,r,dep+1);
    }
    
    
    int query(int L,int R,int l,int r,int dep,int k)
    {
        if(l==r) return tree[dep][l];
        int mid=(L+R)/2;
        int cnt=toleft[dep][r]-toleft[dep][l-1];
        if(cnt>=k)
        {
            int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
            int newr=newl+cnt-1;
            return query(L,mid,newl,newr,dep+1,k);
        }
        else
        {
            int newr=r+toleft[dep][R]-toleft[dep][r];
            int newl=newr-(r-l-cnt);
            return query(mid+1,R,newl,newr,dep+1,k-cnt);
        }
    }
    
    int main()
    {
        int T_T,n,m;
        scanf("%d",&T_T);
        while(T_T--)
        {
            scanf("%d%d",&n,&m);
            for(int i=1;i<=n;i++)
            {
                scanf("%d",sorted+i);
                tree[0][i]=sorted[i];
            }
            sort(sorted+1,sorted+1+n);
            build(1,n,0);
            int l,r,k;
            while(m--)
            {
                scanf("%d%d%d",&l,&r,&k);
                printf("%d
    ",query(1,n,l,r,0,k));
            }
        }
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5139232.html
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