題目鏈接

CF868F. Yet Another Minimization Problem

題解

\(f_{i,j}=\min\limits_{k=1}^{i}\{f_{k,j-1}+w_{k,i}\}\)
\(w_{l,r}\)為區間\([l,r]\)的花費,1D1D的經典形式
發現這個這是個具有決策單調性的轉移
單無法快速轉移,我們考慮分治
對於當前分治區間\([l,r]\) ,它的最優決策區間在\([L,R]\)之間。
對於\([l,r]\)的中點\(mid\)，我們可以暴力掃\([L?mid]\)
找到mid的最優決策點p。因為決策單調，所以\([l,mid?1]\)最優決策區間為\([L,p]\)

決策區間：\(L\frac{\qquad\qquad\qquad\downarrow^{p}\qquad\qquad\qquad}{}R\)

代碼

#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;   
} 
 

CF868 F. Yet Another Minimization Problem 決策單調優化 分治