CF868 F. Yet Another Minimization Problem 决策单调优化 分治

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[TOC]

#题目链接
CF868F. Yet Another Minimization Problem
#题解
\(f_{i,j}=\min\limits_{k=1}^{i}\{f_{k,j-1}+w_{k,i}\}\)
$w_{l,r}\(为区间\)[l,r]$的花费,1D1D的经典形式
发现这个这是个具有决策单调性的转移
单无法快速转移,我们考虑分治
对于当前分治区间$[l,r]$ ,它的最优决策区间在$[L,R]$之间。
对于$[l,r]$的中点$mid$，我们可以暴力扫$[L−mid]$
找到mid的最优决策点p。因为决策单调，所以$[l,mid−1]\(最优决策区间为\)[L,p]\(，而\)[mid+1,r]\(,的最优决策区间在\)[p,R]$上
分治下去
求解区间：\(|\gets预处理\to | l\frac{\qquad\qquad\qquad\downarrow^{mid}\qquad\qquad\qquad}{}r\)

决策区间：\(L\frac{\qquad\qquad\qquad\downarrow^{p}\qquad\qquad\qquad}{}R\)
#代码

#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define LL long long 
#define gc getchar() 
#define pc putchar
inline int read() { 
	int x = 0,f = 1; 
	char c = gc; 
	while(c < '0' || c > '9' )c = gc; 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc; 
	return x * f ;
} 
void print(LL x) {
	 if(x >= 10) print(x / 10); 
	 pc(x % 10 + '0'); 
} 
int n,K; 
const int maxn = 200007; 
int a[maxn],b[maxn],c[maxn]; 
LL f[maxn],dp[maxn]; 
void solve(int l,int r ,int L,int R,int w) {
	if(l > r) return ; 
	int mid = l + r >> 1,k = 0,p = std::min(mid,R); 
	for(int i = l;i <= mid;++ i) w += c[a[i]] ++; 
	for(int i = L;i <= p;++ i) { 
		w -= -- c[a[i]]; 
		if(dp[mid] > f[i] + w) dp[mid] = f[i] + w,k = i; 
	} 
	for(int i = L;i <= p;++ i) w += c[a[i]] ++; 
	for(int i = l;i <= mid;++ i) w -= --c[a[i]]; 
	solve(l,mid - 1,L,k,w); 
	
	for(int i = l;i <= mid;++ i) w += c[a[i]] ++; 
	for(int i = L;i <  k;++ i) w -= -- c[a[i]]; 
	solve(mid + 1,r,k,R,w); 
	
	
	for(int i = L;i < k;++ i) ++ c[a[i]]; 
	for(int i = l;i <= mid;++ i) -- c[a[i]]; 
} 
int main() { 
	n = read(),K = read(); 
	for(int i = 1;i <= n;++ i) 
		f[i] = f[i - 1] + c[a[i] = read()] ++; 
	memset(c,0,sizeof c); 
	for(int i = 1;i <= K;++ i) { 
		memset(dp,0x3f,sizeof dp); 
		solve(1,n,1,n,0); 
		std::swap(f,dp); 
	} 
	print(dp[n]); 
	return 0; 	
}