There are nn students at your university. The programming skill of the ii-th student is aiai. 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⋅1052⋅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 kk students with programming skills a[i1],a[i2],…,a[ik]a[i1],a[i2],…,a[ik], then the diversity of this team is maxj=1ka[ij]−minj=1ka[ij]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.
The first line of the input contains one integer nn (3≤n≤2⋅1053≤n≤2⋅105) — the number of students.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the programming skill of the ii-th student.
In the first line print two integers resres and kk — the minimum total diversity of the division of students and the number of teams in your division, correspondingly.
In the second line print nn integers t1,t2,…,tnt1,t2,…,tn (1≤ti≤k1≤ti≤k), where titi is the number of team to which the ii-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.
5
1 1 3 4 2
3 1
1 1 1 1 1
6
1 5 12 13 2 15
7 2
2 2 1 1 2 1
10
1 2 5 129 185 581 1041 1909 1580 8150
7486 3
3 3 3 2 2 2 2 1 1 1
In the first example, there is only one team with skills [1,1,2,3,4][1,1,2,3,4] so the answer is 33. It can be shown that you cannot achieve a better answer.
In the second example, there are two teams with skills [1,2,5][1,2,5] and [12,13,15][12,13,15] so the answer is 4+3=74+3=7.
In the third example, there are three teams with skills [1,2,5][1,2,5], [129,185,581,1041][129,185,581,1041] and [1580,1909,8150][1580,1909,8150] so the answer is 4+912+6570=74864+912+6570=7486.
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> using namespace std; #define mem(s,t) memset(s,t,sizeof(s)) #define pq priority_queue #define pb push_back #define fi first #define se second #define ac return 0; #define ll long long #define cin2(a,n,m) for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j]; #define rep_(n,m) for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) #define rep(n) for(int i=1;i<=n;i++) #define test(xxx) cout<<" Test " <<" "<<xxx<<endl; #define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10); #define lc now<<1 #define rc now<<1|1 #define ls now<<1,l,mid #define rs now<<1|1,mid+1,r #define half no[now].l+((no[now].r-no[now].l)>>1) #define ll long long #define inf 0x3f3f3f3f const int mxn = 2e5+10; ll n,m,k,ans,cnt,col; int dp[mxn],pro[mxn],flag[mxn];//dp[] 最小化 pro[] 下标 flag 分组记录 string str,ch ; pair <int,int> mp[mxn]; bool cmp(pair<int,int>x,pair<int,int>y) {return x.first>y.first;} int main() { while(cin>>n) { for(int i=1;i<=n;i++) { cin>>mp[i].first , mp[i].second=i; dp[i] = inf; } sort(mp+1,mp+1+n,cmp); dp[0] = 0; for(int i=3;i<=n;i++) { if(i-3>=0 && dp[i]>dp[i-3]+mp[i-2].first-mp[i].first) { dp[i] = dp[i-3]+mp[i-2].first-mp[i].first; pro[i] = i-3; } if(i-4>=0 && dp[i]>dp[i-4]+mp[i-3].first-mp[i].first) { dp[i] = dp[i-4]+mp[i-3].first-mp[i].first; pro[i] = i-4; } if(i-5>=0 && dp[i]>dp[i-5]+mp[i-4].first-mp[i].first) { dp[i] = dp[i-5]+mp[i-4].first-mp[i].first; pro[i] = i-5; } } ans = pro[n] , col = n,cnt = 1; while(col) { while(col>ans) { flag[mp[col].second] = cnt; col--; } col = ans ; ans = pro[col]; cnt++; } cout<<dp[n]<<" "<<--cnt<<endl; for(int i=1;i<=n;i++) cout<<flag[i]<<" "; cout<<endl; } return 0; }
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define mem(s,t) memset(s,t,sizeof(s))
#define pq priority_queue
#define pb push_back
#define fi first
#define se second
#define ac return 0;
#define ll long long
#define cin2(a,n,m) for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j];
#define rep_(n,m) for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)
#define rep(n) for(int i=1;i<=n;i++)
#define test(xxx) cout<<" Test " <<" "<<xxx<<endl;
#define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10);
#define lc now<<1
#define rc now<<1|1
#define ls now<<1,l,mid
#define rs now<<1|1,mid+1,r
#define half no[now].l+((no[now].r-no[now].l)>>1)
#define ll long long
#define inf 0x3f3f3f3f
const int mxn = 2e5+10;
ll n,m,k,ans,cnt,col;
int dp[mxn],pro[mxn],flag[mxn];
string str,ch ;
pair <int,int> mp[mxn];
int main()
{
while(cin>>n)
{
memset(dp,inf,sizeof(dp));
for(int i=1;i<=n;i++)
cin>>mp[i].first , mp[i].second=i;
sort(mp+1,mp+1+n); dp[1] = 0;
for(int i=1;i<=n;i++)
{
for(int k=2;k<=4 && i+k<=n;k++)
{
int ans = mp[i+k].first-mp[i].first; //计算区间差值
if(dp[i+k+1]>dp[i]+ans)
{
dp[i+k+1] = dp[i]+ans;
pro[i+k+1] = i ; //存储方案 i ~ i+k 为一组
}
}
}
cnt = 1 , col = n+1;
while(col!=1)
{
for(int i=col-1;i>=pro[col];i--)
{
flag[mp[i].second] = cnt;
}
cnt++;
col=pro[col];
}
cout<<dp[n+1]<<" "<<--cnt<<endl;
for(int i=1;i<=n;i++)
cout<<flag[i]<<" ";
cout<<endl;
}
return 0;
}