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;
}