Codeforces Round #611 (Div. 3) D

Posted 2020-11-25 lipoicyclic

There are nn Christmas trees on an infinite number line. The ii -th tree grows at the position xixi . All xixi are guaranteed to be distinct.

Each integer point can be either occupied by the Christmas tree, by the human or not occupied at all. Non-integer points cannot be occupied by anything.

There are mm people who want to celebrate Christmas. Let y1,y2,…,ymy1,y2,…,ym be the positions of people (note that all values x1,x2,…,xn,y1,y2,…,ymx1,x2,…,xn,y1,y2,…,ym should be distinct and all yjyj should be integer). You want to find such an arrangement of people that the value ∑j=1mmini=1n|xi−yj|∑j=1mmini=1n|xi−yj| is the minimum possible (in other words, the sum of distances to the nearest Christmas tree for all people should be minimized).

In other words, let djdj be the distance from the jj -th human to the nearest Christmas tree (dj=mini=1n|yj−xi|dj=mini=1n|yj−xi| ). Then you need to choose such positions y1,y2,…,ymy1,y2,…,ym that ∑j=1mdj∑j=1mdj is the minimum possible.

Input

The first line of the input contains two integers nn and mm (1≤n,m≤2⋅1051≤n,m≤2⋅105 ) — the number of Christmas trees and the number of people.

The second line of the input contains nn integers x1,x2,…,xnx1,x2,…,xn (−109≤xi≤109−109≤xi≤109 ), where xixi is the position of the ii -th Christmas tree. It is guaranteed that all xixi are distinct.

Output

In the first line print one integer resres — the minimum possible value of ∑j=1mmini=1n|xi−yj|∑j=1mmini=1n|xi−yj| (in other words, the sum of distances to the nearest Christmas tree for all people).

In the second line print mm integers y1,y2,…,ymy1,y2,…,ym (−2⋅109≤yj≤2⋅109−2⋅109≤yj≤2⋅109 ), where yjyj is the position of the jj -th human. All yjyj should be distinct and all values x1,x2,…,xn,y1,y2,…,ymx1,x2,…,xn,y1,y2,…,ym should be distinct.

If there are multiple answers, print any of them.

一开始理解错题面了，用中位数思路写了一通发现连样例都过不了。这个题的意思是确定n个的位置(人和人，人和树的位置不能有重合)，使每个人距离最近的圣诞树的距离的和最小。首先想到把人安排在圣诞树两边(人的位置为树的位置+-1)，两边如果被占据的话继续往两边移。因为题目求的是和最小，容易想到bfs的思想(搜到终点能保证最小),所以在此处利用bfs，并利用map来存储当前位置到最近的树的距离。首先把所有树的坐标扔进队列。从队列取数后，先判断这个位置是否为空位置：if(d[temp]!=0) 为空的话（即字典里没有过这个键对）直接扔进存最终答案的vector里，更新总距离。然后判断这个位置加减1后如果仍然没出现在字典里，把这个位置的距离更新，扔进队列里。循环结束的条件是vector的大小等于m。最终输出就可以了。

(第一次写博客，写的不好还请见谅

 

#include<bits/stdc++.h>
using namespace std;
int n,m;
int x[200005]={0};
 vector<int>v;
 queue<int>q;
 map<int,int>d;
int main()
{
 cin>>n>>m;
 int i;
 for(i=1;i<=n;i++)
 {
  scanf("%d",&x[i]);
  q.push(x[i]);
  d[x[i]]=0;
 }
 long long tot_dis=0;
 int cnt=0;
 while(!q.empty())
 {
  if(v.size()>=m)break;
  int temp=q.front();
  q.pop();
  if(d[temp]!=0)
  {
   tot_dis+=d[temp];
   v.push_back(temp);
  }
  if(d.count(temp-1)==0)
  {
   d[temp-1]=d[temp]+1;
   q.push(temp-1);
  }
  if(d.count(temp+1)==0)
  {
   d[temp+1]=d[temp]+1;
   q.push(temp+1);
  }
  
 }
 cout<<tot_dis<<endl;
 int j;
 for(j=0;j<v.size();j++)
 {
  cout<<v[j]<<‘ ‘;
 }
 return 0;
}