codeforces 1283D BFS

太久没写bfs，写法想错了算错了复杂度

题意：x坐标上，给了n个圣诞树的坐标，让你再求出m个人点，使得这m个人的点到最近的圣诞树的和最小，这n+m个点distinct。

思路：之间裸bfs。因为x1e9，可以用map<int,int> vis;//用map记录是否访问过。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
map<int,int> vis;//用map记录是否访问过
ll a[maxn];
int n,m;
ll sum = 0;
priority_queue<pair<ll,ll> > que;
queue<ll>ans;
void solve()
{
    int cnt = 0;
    while(que.size()&&ans.size()<m)
    {

        int d = que.top().fi;
        d = -d;
        int pos = que.top().se;
        que.pop();
        if(vis[pos]==1) continue;
        vis[pos]=1;
        ans.push(pos);
        sum += d;
        que.push(mp(-(d+1),pos-1));
        que.push(mp(-(d+1),pos+1));
    }
}
int main()
{
    //freopen("input.txt", "r", stdin);
    cin>>n>>m;
    rep(i,1,n){
        cin>>a[i];
        vis[a[i]]=1;
    }
    rep(i,1,n){
        que.push(mp(-1,a[i]-1));
        que.push(mp(-1,a[i]+1));
    }
    solve();
    cout<<sum<<endl;
    while(ans.size()){
        int x = ans.front();
        ans.pop();
        cout<<x<<sp;
    }



}