CF868F Yet Another Minimization Problem

You are given an array of nn integers a_{1}...\ a_{n}a1?... an? . The cost of a subsegment is the number of unordered pairs of distinct indices within the subsegment that contain equal elements. Split the given array into kk non-intersecting non-empty subsegments so that the sum of their costs is minimum possible. Each element should be present in exactly one subsegment.

The first line contains two integers nn and kk ( 2<=n<=10^{5}2<=n<=105 , 2<=k<=min\ (n,20))2<=k<=min (n,20)) — the length of the array and the number of segments you need to split the array into.

The next line contains nn integers a_{1},a_{2},...,a_{n}a1?,a2?,...,an? ( 1<=a_{i}<=n1<=ai?<=n ) — the elements of the array.

Print single integer: the minimum possible total cost of resulting subsegments.

题目描述：给定一个序列 aa，要把它分成 kk 个子段。每个子段的费用是其中相同元素的对数。求所有子段的费用之和的最小值。

输入格式：第一行输入 nn（序列长度）和 kk（需分子段数）。接下来一行有 nn 个数，第 ii 个数表示序列的第 ii 个元素 a_iai?。

输出格式：输出一个数，费用和的最小值。

2 \leq n \leq 10^52≤n≤105，2 \leq k \leq min(n,20)2≤k≤min(n,20)，1 \leq a_i \leq n1≤ai?≤n 。

In the first example it‘s optimal to split the sequence into the following three subsegments: [1][1] , [1,3][1,3] , [3,3,2,1][3,3,2,1] . The costs are 00 , 00 and 11 , thus the answer is 11 .

In the second example it‘s optimal to split the sequence in two equal halves. The cost for each half is 44 .

In the third example it‘s optimal to split the sequence in the following way: [1,2,2,2,1][1,2,2,2,1] , [2,1,1,1,2][2,1,1,1,2] , [2,1,1][2,1,1] . The costs are 44 , 44 , 11 .

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int c[100005],tr,tl,f[30][100005],a[100005],s,n,k;
 7 int get(int l,int r)
 8 {
 9     while (tl<l) 
10     {
11         c[a[tl]]--;
12         s-=c[a[tl]];
13         tl++;
14     }
15     while (tl>l)
16     {
17         tl--;
18         s+=c[a[tl]];
19         c[a[tl]]++;
20     }
21     while (tr<r)
22     {
23         tr++;
24         s+=c[a[tr]];
25         c[a[tr]]++;
26     }
27     while (tr>r)
28     {    
29         c[a[tr]]--;
30         s-=c[a[tr]];
31         tr--;
32     } 
33     return s;
34 }
35 void solve(int x,int l,int r,int L,int R)
36 {int i,Mid;
37     if (l>r) return;
38     int mid=(l+r)/2;
39     Mid=mid;
40     for (i=L;i<=min(mid-1,R);i++)
41     {
42         int t=f[x-1][i]+get(i+1,mid);
43         if (f[x][mid]>=t) 
44         {
45             f[x][mid]=t;
46             Mid=i;
47         }
48     }
49     solve(x,l,mid-1,L,Mid);
50     solve(x,mid+1,r,Mid,R);
51 }
52 int main()
53 {int i,j;
54     scanf("%d%d",&n,&k);
55     for (i=1;i<=n;i++)
56     scanf("%d",&a[i]);
57     memset(f,127/2,sizeof(f));
58     tl=1;tr=0;
59     for (i=1;i<=n;i++)
60     {
61         f[1][i]=get(1,i);
62     } 
63     for (i=2;i<=k;i++)
64     {
65         solve(i,i,n,i-1,n);
66     }
67     printf("%d",f[k][n]);
68 }