Codeforces Round #598 (Div. 3) E. Yet Another Division Into Teams dp

E. Yet Another Division Into Teams

There are n students at your university. The programming skill of the i-th student is ai. As a coach, you want to divide them into teams to prepare them for the upcoming ICPC finals. Just imagine how good this university is if it has 2⋅105 students ready for the finals!

Each team should consist of at least three students. Each student should belong to exactly one team. The diversity of a team is the difference between the maximum programming skill of some student that belongs to this team and the minimum programming skill of some student that belongs to this team (in other words, if the team consists of k students with programming skills a[i1],a[i2],…,a[ik], then the diversity of this team is maxj=1ka[ij]−minj=1ka[ij]).

The total diversity is the sum of diversities of all teams formed.

Your task is to minimize the total diversity of the division of students and find the optimal way to divide the students.

Input

The first line of the input contains one integer n (3≤n≤2⋅105) — the number of students.

The second line of the input contains n integers a1,a2,…,an (1≤ai≤109), where ai is the programming skill of the i-th student.

Output

In the first line print two integers res and k — the minimum total diversity of the division of students and the number of teams in your division, correspondingly.

In the second line print n integers t1,t2,…,tn (1≤ti≤k), where ti is the number of team to which the i-th student belong.

If there are multiple answers, you can print any. Note that you don't need to minimize the number of teams. Each team should consist of at least three students.

Examples

input
5
1 1 3 4 2
output
3 1
1 1 1 1 1
input
6
1 5 12 13 2 15
output
7 2
2 2 1 1 2 1
input
10
1 2 5 129 185 581 1041 1909 1580 8150
output
7486 3
3 3 3 2 2 2 2 1 1 1

Note

In the first example, there is only one team with skills [1,1,2,3,4] so the answer is 3. It can be shown that you cannot achieve a better answer.

In the second example, there are two teams with skills [1,2,5] and [12,13,15] so the answer is 4+3=7.

In the third example, there are three teams with skills [1,2,5], [129,185,581,1041] and [1580,1909,8150] so the answer is 4+912+6570=7486.

题意

这个学校里面有n个学生，你须要给他们分红若干的队伍，每一个队伍最少3我的。

每一个队伍定义差别值是这个队伍最强的人和最弱的人的能力值差。

如今你须要构建若干个队伍，使得差别值的总和最小。

题解

咱们先排序，那么分队伍必定是选择排序后的连续几我的组成一队。

而后每一个队伍必定人数最多为5我的，由于6我的就能够拆成两队，而后两队的代价必定是比一个队伍的代价小。

而后就是个简单的dp了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n;
pair<int,int> k[maxn];
int dp[maxn];
int p[maxn];
int fr[maxn];
int ans_pos[maxn];
int tot=0;
void dfs(int x){
    if(x==0)return;
    tot++;
    p[x]=1;
    dfs(fr[x]);
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&k[i].first);
        k[i].second=i;
    }
    sort(k+1,k+1+n);
    memset(dp,-1,sizeof(dp));
    dp[0]=0;
    dp[3]=k[3].first-k[1].first;
    for(int i=4;i<=n;i++){
        for(int j=3;j<=6;j++){
            if(dp[i-j]!=-1){
                if(dp[i]==-1){
                    dp[i]=dp[i-j]+k[i].first-k[i-j+1].first;
                    fr[i]=i-j;
                }
                else{
                    if(dp[i-j]+(k[i].first-k[i-j+1].first)<dp[i]){
                        fr[i]=i-j;
                        dp[i]=dp[i-j]+k[i].first-k[i-j+1].first;
                    }
                }
            }
        }
    }

    dfs(n);
    cout<<dp[n]<<" "<<tot<<endl;
    int tot2=1;
    for(int i=1;i<=n;i++){
        if(p[i]==0){
            p[i]=tot2;
        }else if(p[i]==1){
            p[i]=tot2;
            tot2++;
        }
    }
    for(int i=1;i<=n;i++){
        ans_pos[k[i].second]=p[i];
    }
    for(int i=1;i<=n;i++){
        cout<<ans_pos[i]<<" ";
    }
    cout<<endl;
}