Hdu 2665 Kth number

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2301    Accepted Submission(s): 782

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

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

Recommend

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 划分树

时间复杂度O( nlog(n) + mlog(n) )     n代表数据元素个数，m代表操作次数

空间复杂度O( nlog(n))

#include<stdio.h>
#include<algorithm>
using namespace std;

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define MAXN 110000

int sorted[MAXN];
int val[20][MAXN];
int toL[20][MAXN];
int n,m;

void build(int l,int r,int rt,int d)
{
    if(l==r)
    {
        return ;
    }
    int i,count,mid=(l+r)>>1;
    int tmp=sorted[mid];
    count=mid-l+1;
    for(i=l;i<=r;i++)
    {
        if(val[d][i]<tmp)
            count--;
    }
    int lpos=l,rpos=mid+1;
    toL[d][0]=0;
    for(i=l;i<=r;i++)
    {
        if(val[d][i]<tmp)
        {
            val[d+1][lpos++]=val[d][i];
            toL[d][i]=toL[d][i-1]+1;
        }
        else if(val[d][i]>tmp)
        {
            val[d+1][rpos++]=val[d][i];
            toL[d][i]=toL[d][i-1];
        }
        else if(count>0)
        {
            val[d+1][lpos++]=val[d][i];
            toL[d][i]=toL[d][i-1]+1;
            count--;
        }
        else
        {
            val[d+1][rpos++]=val[d][i];
            toL[d][i]=toL[d][i-1];
        }

    }
    build(lson,d+1);
    build(rson,d+1);
}

int query(int L,int R,int k,int l,int r,int rt,int d)
{
    if(l==r)
        return val[d][L];
    int s1;
    int s2;
    if(L!=l)
    {
        s1=toL[d][R]-toL[d][L-1];
        s2=toL[d][L-1]-toL[d][l-1];
    }
    else
    {
        s1=toL[d][R]-toL[d][L-1];
        s2=0;
    }
    int new_L,new_R;
    int mid=(l+r)>>1;
    if(k<=s1)
    {
        new_L=l+s2;
        new_R=l+s1+s2-1;
        return query(new_L,new_R,k,lson,d+1);
    }
    else
    {
        k=k-s1;
        s1=(R-L+1)-s1;
        s2=L-l-s2;
        new_L=(mid+1)+s2;
        new_R=(mid+1)+s1+s2-1;
        return query(new_L,new_R,k,rson,d+1);
    }

}
int main(void)
{
//    freopen("d:\\in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int i;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[0][i]);
            sorted[i]=val[0][i];
        }
        sort(sorted+1,sorted+1+n);
        build(1,n,1,0);
        int a,b,k;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&k);
            printf("%d\n",query(a,b,k,1,n,1,0));
        }
    }
    return 0;
}