CF868F Yet Another Minimization Problem 解题报告

题意翻译

题目描述：给定一个序列 $a$，要把它分成 $k$ 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 $n$（序列长度）和 $k$（需分子段数）。接下来一行有 $n$ 个数，第 $i$ 个数表示序列的第 $i$ 个元素 $a_i$。 输出格式：输出一个数，费用和的最小值。 $2 \leq n \leq 10^5$，$2 \leq k \leq min(n,20)$，$1 \leq a_i \leq n$ 。

题目描述

You are given an array of $n$ integers $a_{1}...\ a_{n}$ . 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 $k$ non-intersecting non-empty subsegments so that the sum of their costs is minimum possible. Each element should be present in exactly one subsegment.

输入输出格式

输入格式

The first line contains two integers $n$ and $k$ ( $2<=n<=10^{5}$ , $2<=k<=min\ (n,20))$ — the length of the array and the number of segments you need to split the array into. The next line contains $n$ integers $a_{1},a_{2},...,a_{n}$ ( $1<=a_{i}<=n$ ) — the elements of the array.

输出格式

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

输入输出样例

输入样例 #1

7 3
1 1 3 3 3 2 1

输出样例 #1

1

输入样例 #2

10 2
1 2 1 2 1 2 1 2 1 2

输出样例 #2

8

输入样例 #3

13 3
1 2 2 2 1 2 1 1 1 2 2 1 1

输出样例 #3

9

说明

In the first example it's optimal to split the sequence into the following three subsegments: $[1]$ , $[1,3]$ , $[3,3,2,1]$ . The costs are $0$ , $0$ and $1$ , thus the answer is $1$ . In the second example it's optimal to split the sequence in two equal halves. The cost for each half is $4$ . In the third example it's optimal to split the sequence in the following way: $[1,2,2,2,1]$ , $[2,1,1,1,2]$ , $[2,1,1]$ . The costs are $4$ , $4$ , $1$ .

SOLUTION

首先来一波BF~ 🤢

设$f_{i,j}$表示前$i$个数分成$j$段的最小费用

$$f_{i,j}=min_{k<i}(f_{k,j-1}+w(k+1,i))$$

emmm...很好，$O(kn^2)$，直接寄掉。

于是我们考虑优化

我们可以浅浅发现好像这个dp满足决策单调性

因为$f_{i,j}$的最优决策点肯定小于等于$f_{i+1,j}$

具体证明：（略）

于是我们就可以通过分治优化dp

接下来的问题就是如何求出$w(k+1,i)$了

我们可以借助莫队的思想，挪动l,r端点来达到优化时间的目的

总复杂度$O(knlogn)$

CODE

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+2;
ll L=1,R,f[N][23],a[N],cnt[N],n,m,ans,k;
ll w(ll l, ll r){//动态算一下贡献,类似于莫队 
	while(L<l) ans-=(--cnt[a[L++]]);
	while(R>r) ans-=(--cnt[a[R--]]);
	while(L>l) ans+=cnt[a[--L]]++;
	while(R<r) ans+=cnt[a[++R]]++;
	return ans;
}
void solve(ll l,ll r,ll L,ll R){//分治 
	if(l>r) return;
	ll p=0,mid=(l+r)>>1;
	for(ll i=L; i<=min(R,mid-1); i++){//p:最优决策点
		if(f[i][k-1]+w(i+1,mid)<f[mid][k]) f[mid][k]=f[p=i][k-1]+w(i+1,mid);
	}
	solve(l,mid-1,L,p),solve(mid+1,r,p,R);
}
int main(){
	scanf("%lld%lld",&n,&m);
	for(ll i=1; i<=n;i++){
		scanf("%lld",&a[i]);
	}
	memset(f,0x3f,sizeof f);
	for(ll i=1; i<=n; i++){
		f[i][1]=w(1,i);
	}
	for(ll i=2; i<=m; i++){
		k=i,solve(1,n,0,n-1);
	}
	printf("%lld",f[n][m]);
	return 0;
}

完结撒花❀

★,°:.☆(￣▽￣)/$:.°★ 。