"8.20 simulation" bubble sort

Title Description

given the n,k, and a sequence of length n, output the result of this sequence after the bubble sort K times.

Thinking of solving problems

We look at the pseudo-code given above, and we can see that this is a code that sorts the code into ascending order, so let's consider some of the characteristics of a counterfeit sort.

A large number to the right to exchange, but once the exchange can be changed a lot of positions, so do not know where to go to change once, so it is difficult to maintain.

A small number is exchanged at most and the larger number of its left one at a time, so at most once a position is run to the left, which is not known better than the large number of maintenance to go there.

Further thinking, after the K-Exchange, up to the left to run the K-step, that is to say, the number of k+1 position can run to 1 of the position up, then we can boldly guess, the first k+1 number of the number of the second after the exchange of the smallest will be the first position. If the smallest number of former k+1 does not run to the first position, it means that there must be a smaller number than it will be in its left to block it, and our assumptions contradict, so conjecture is established.

And so on, the number of k+2 is up to the second position, and according to our conclusion, the second position should be the smallest number except the number of the first position between 1 ~ k+2.

Then push down and we're done with all the numbers in the position of the calculation.

As can be seen, we need to maintain a minimum value, obviously can use the line segment tree, you can use the priority queue, note that the push to use a negative value.

Code

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <queue>6 using namespacestd;7 Const intmaxn=100050;8 intn,k;9 intA[MAXN];Tenpriority_queue<int>Q; One intMain () { Ascanf"%d%d",&n,&k); -      for(RegisterintI=1; i<=n;i++) scanf ("%d",&a[i]); -      for(RegisterintI=1; I<=min (n,k+1); i++){ theQ.push (-a[i]); -     } -      for(RegisterintI=2; i+k<=n;i++){ -printf"%d\n",-q.top ()); + Q.pop (); -Q.push (-a[i+K]); +     } A      while(!Q.empty ()) { atprintf"%d\n",-q.top ()); - Q.pop (); -     } -}

"8.20 simulation" bubble sort