Blue Bridge simulation - Question 10 - party playbill

title: Blue bridge simulation _ _ Q10 party playbill
categories:

ACM
RMQ
tags:
ST
date: 2020-03-27 18:20:44

The most frequent queries value for the interval is traversed overhead is very large, it produces a range of best value queries this algorithm, we talk about the ST + RMQ, ST is sparse mean, this approach requires data must be static It can not change.

Issues into

There is a one array (of length n), frequent queries [s, e] most value interval.

Detailed algorithms

Define a two-dimensional array ST (size [n, lgn]). ST[i][j]I represents the beginning from (including i) rearwardly 2 ^{J-th element value most section, i.e. the interval [i, i + 2} J-. 1] most value, then it is obvious ST[i][0]=a[i](because [i, i] value is only a ). Now we start recursive derivation, [I, I + 2 ^{J-. 1] can be divided into two according to the size of the intermediate 2} (. 1-J) interval [I, I + 2 ^{(. 1-J) -1] and [I + 2} (-J. 1), I + 2 ^{J-. 1], then [I, I + 2} J-. 1] value of the two most sections then most value, i.e. [I, I + 2 ^{J -1] = min / max ([I, I 2 +} (. 1-J) -1], [I 2 + ^{(. 1-J), I 2 +} J-. 1]) is:

ST[i][j]=min/max(ST[i][j-1],ST[i+2^j][j-1])

Then you can find an element to calculate the value of the j-th column is saying only two values of a previous column about, and we have just considered a good first row, first column can be calculated by the second column, the second column The third column can be calculated to count ...... so where one does? ST[i][j]It represents the interval [I, I + 2 ^{J-. 1] + 2 obviously I} J-. 1 <=. 1-n-(n-elements, the subscript 0 ~ n-1) so there is a built templates ST:

void cal(int a[],int n)
{
	for(int i=0;i<=n-1;i++)
		ST[i][0]=a[i];
	for(int j=1;(1<<(j))<=n;j++)
	{
		for(int i=0;i+(1<<j)-1<=n-1;i++)
		{
			st[i][j]=max/min(st[i][j-1],st[i+1<<(j-1)][j-1]);
		}
	}	
}

With this table for a long time to say that how to use it? If we want to calculate the interval [s, e] the best value for how to count it? Is that we can use ST[此处可为任意数][此处必须为2的幂]if we can find a k, so that the interval [s, e] into [S, S + 2 ^{K-. 1] and `[s + 2 ^ k] [e]`, we can use the table , but you will not find a second interval that we can only express the length of the interval is a power of 2. So smart people will find that using e minus the power of two is not on the line! Thus the [s, e] into [S, S + 2} K-. 1] and [e-2 ^ k + 1 , e], so that the interval [s][e]most value can be changed min/max(ST[s][k],ST[e-2^K+1][e])but it was possible to turn the question, If the intersection of the two sets do not how to do? For example [2,6] into [2,3] and [5,6] (k = 1) , thus obtained, then obviously not ah! So we need to find a relatively large value of k, [S, 2 S + ^{k-. 1] must be included in the `[s] [e]` inside, i.e. 2 + S} k. 1-<solve for E =
$k<=log_2(e-s+1),令k=log_2(e-s+1)$
and (int) removes the fractional portion k of k, that is 0 <= k- (int) kso the k-1 <(int) ksubstituting the original formula:
$2 ^ {k-1} -1 <2 ^ {(int) k} -1 <= 2 ^ k-1 to k = log_2 (e-s + 1) is substituted to give$

$(Es) / 2-1 / 2 <2 ^ {(int) k} -1, where 2 ^ {(int) k} -1 is an integer, (es) / 2 = <2 ^ {(int) k} -1$

$Therefore, s + 2 ^ {(int) k} -1> = s + (es) / 2 = (e + s) / 2 (midpoint) and e-2 ^ k + 1 <= e- (e + s) / 2 = (e + s) / 2 (midpoint)$

So when k = (int) (log2 ( e-s + 1)) , the ST[s][k],ST[e-2^K+1][e]interval must be indicated by the intersection. So in claim [s, e] the best values requires calculation intervalmin/max(ST[s][k],ST[e-2^K+1][e]),k=(int)(log2(e-s+1))

example

Q10 party playbill

topic

【Problem Description】

Xiao Ming to organize a party, prepared a total of n program. Then the evening time is limited, he can only chose one of the m program.
This program is in the order of the n Xiaoming contemplated given the order can not be changed.
Xiao Ming found that the audience likes the degree of the party has a very large extent the relationship and look good the first few shows, he hopes the first program selected as good-looking, in this context you want the second program as good-looking, in turn analogy.
Xiao Ming to each program defines a nice value, you help Xiao Ming selected m programs, meet his demands.

[Input Format]

The first line of input contains two integers n, m, and the number represents the number of a program to be selected.
The second line contains n integers, followed Attractive value for each program.

[Output format]

M output line contains integers, looking for the value of the selected program.

[Sample input]

5 3
3 1 2 5 4

[Sample output]

3 5 4

[Sample Description]

Selecting the first, 4, 5 programs.

[Evaluation] Examples scale and agreed with

For 30% of the evaluation use cases, 1 <= n <= 20 ;
for 60% of the evaluation use cases, 1 <= n <= 100 ;
for all reviews use cases, 1 <= n <= Attractive value 100000,0 <= program <= 100,000.

Thinking O (N ^ 2)

ST+RMQ

Reference Code

#include<cstdio>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<iostream>
#include<stdlib.h>
#define PP(a,b) cout<<a<<"="<<b<<endl
#define P(a) cout<<a<<endl
#define print(a) cout<<#a<<"="<<a<<endl
#define _for(i,s,e) for(int i=s;i<=e;i++)
using namespace std;
int n,m;
int a[100001];
int st[100001][20];
void cal(int a[],int n)
{
	for(int j=1;(1<<(j-1))<=n;j++)
	{
		for(int i=0;i+(1<<j)-1<=n-1;i++)
		{
			st[i][j]=max(st[i][j-1],st[i+1<<(j-1)][j-1]);
		}
	}	
}
int query(int s,int e)
{
	int t=(int)(log(e-s+1)/log(2));
	return max(st[s][t],st[e-(1<<t)+1][t]);
} 
int main()
{	
	//freopen("input.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m;
    _for(i,0,n-1)
    {
    	cin>>a[i];
    	st[i][0]=a[i];
    }
    cal(a,n);
    int s=0,k=m;
    while(k!=0)
    {
    	int z=query(s,n-1-k+1);
    	cout<<z<<" ";
    	s=z+1;
    	k--;
    }        	  
}