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

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;   
}