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

There are

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

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

The second line of the input contains

Output

In the first line print two integers

In the second line print

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

Copy

5
1 1 3 4 2

output

Copy

3 1
1 1 1 1 1

input

Copy

6
1 5 12 13 2 15

output

Copy

7 2
2 2 1 1 2 1

input

Copy

10
1 2 5 129 185 581 1041 1909 1580 8150

output

Copy

7486 3
3 3 3 2 2 2 2 1 1 1

Note

In the first example, there is only one team with skills

In the second example, there are two teams with skills

In the third example, there are three teams with skills

/*大概意思： 这个学校里面有n个学生，你需要给他们分成若干的队伍，每个队伍最少3个人。
每个队伍定义差异值是这个队伍最强的人和最弱的人的能力值差。
现在你需要构建若干个队伍，使得差异值的总和最小。*/
/* 大概思路：先排序，然后选择连续的几个人排成一队
每个队为最少3个，最多五个因为6个人就可以拆成两队，两队的差值一定比一个队伍的小
然后 dp*/
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n,tot=0,ans_pos[maxn],fr[maxn],p[maxn],dp[maxn];
pair<int,int> k[maxn];
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;
}

原文地址：https://www.cnblogs.com/QingyuYYYYY/p/11804168.html