hdu 2665 Kth number 划分树

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2354    Accepted Submission(s): 800

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

zty

//上次写的POJ kth number是每个数不同的
//而这次数可能重复，所以要修改下策略，感觉这个更好更强大

#include <iostream>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

#define N 100002

#define lson l,m,lv+1

#define rson m+1,r,lv+1

int tr[22][N];

int num[22][N];

int st[N];

int ar[N];

void build(int l,int r,int lv)

{

    if(l==r) return;

    int m=(l+r)>>1;

    int same=0,i;

    for(i=m;i>=l;i--)

      if(st[i]==st[m])

        same++;

      else

        break;

    int lid=l,rid=m+1;

    for(i=l;i<=r;i++)

     if(tr[lv][i]<st[m])

     {

         num[lv][i]=i==l?1:num[lv][i-1]+1;

         tr[lv+1][lid++]=tr[lv][i];

     }

     else if(tr[lv][i]==st[m]&&same>0)

     {

         num[lv][i]=i==l?1:num[lv][i-1]+1;

         tr[lv+1][lid++]=tr[lv][i];

         same--;

     }

     else

     {

           num[lv][i]=i==l?0:num[lv][i-1];

           tr[lv+1][rid++]=tr[lv][i];

     }

    build(lson);

    build(rson);

}

void query(int L,int R,int k,int l,int r,int lv)//整个这个里面操作都简单了许多、、、

{

      if(l==r)

      {

          printf("%d\n",tr[lv][l]);

          return ;

      }

      int m=(l+r)>>1;

      int cnt=num[lv][l+R-1]-(L==1?0:num[lv][l+L-2]);//求区间 里面第 L个数到 第 R个数有几个被划分到左边

      if(cnt>=k)

      {

          int lid=L==1?1:(num[lv][l+L-2]+1);

          int rid=cnt+lid-1;

          query(lid,rid,k,lson);

      }

      else

      {

          int lid=L==1?1:(L-num[lv][l+L-2]); //建立新的查询区间

          int rid=R-num[lv][l+R-1];

          query(lid,rid,k-cnt,rson);

      }

}

int main()

{

    int T;

    int n,m;

    int i;

    scanf("%d",&T);

    while(T--)

    {

       scanf("%d %d",&n,&m);

       for(i=1;i<=n;i++)

        {

            scanf("%d",&tr[0][i]);

            st[i]=tr[0][i];

        }

       sort(st+1,st+i);

       build(1,n,0);

       int l,r,k;

       for(i=0;i<m;i++)

       {

           scanf("%d %d %d",&l,&r,&k);

           query(l,r,k,1,n,0);

       }

    }

    return 0;

}