CF868F Yet Another Minimization Problem [决策单调性]

\(calc(i,j)\) 同样满足四边形不等式，于是没了。

// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i , x , y) for(register int i = (x) ; i <= (y) ; ++ i)
#define Rep(i , x , y) for(register int i = (x) ; i >= (y) ; -- i)
using namespace std ;
using db = double ;
using ll = long long ;
using uint = unsigned int ;
#define int long long
using pii = pair < int , int > ;
#define ve vector
#define Tp template
#define all(v) v.begin() , v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp < class T > void cmax(T & x , const T & y) { if(x < y) x = y ; }
Tp < class T > void cmin(T & x , const T & y) { if(x > y) x = y ; }
// sort , unique , reverse
Tp < class T > void sort(ve < T > & v) { sort(all(v)) ; }
Tp < class T > void unique(ve < T > & v) { sort(all(v)) ; v.erase(unique(all(v)) , v.end()) ; }
Tp < class T > void reverse(ve < T > & v) { reverse(all(v)) ; }
const int SZ = 0x191981 ;
struct FILEIN {
	~ FILEIN () {} char qwq[SZ] , * S = qwq , * T = qwq , ch ;
	char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq , 1 , SZ , stdin) , S == T) ? EOF : * S ++ ; }
	FILEIN & operator >> (char & c) { while(isspace(c = GETC())) ; return * this ; }
	FILEIN & operator >> (string & s) {
		while(isspace(ch = GETC())) ; s = ch ;
		while(! isspace(ch = GETC())) s += ch ; return * this ;
	}
	Tp < class T > void read(T & x) {
		bool sign = 1 ; while((ch = GETC()) < 0x30) if(ch == 0x2d) sign = 0 ;
		x = (ch ^ 0x30) ; while((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30) ;
		x = sign ? x : -x ;
	}
	FILEIN & operator >> (int & x) { return read(x) , * this ; }
	FILEIN & operator >> (signed & x) { return read(x) , * this ; }
	FILEIN & operator >> (unsigned & x) { return read(x) , * this ; }
} in ;
struct FILEOUT { const static int LIMIT = 0x114514 ;
	char quq[SZ] , ST[0x114] ; signed sz , O ;
	~ FILEOUT () { sz = O = 0 ; }
	void flush() { fwrite(quq , 1 , O , stdout) ; fflush(stdout) ; O = 0 ; }
	FILEOUT & operator << (char c) { return quq[O ++] = c , * this ; }
	FILEOUT & operator << (string str) {
		if(O > LIMIT) flush() ; for(char c : str) quq[O ++] = c ; return * this ;
	}
	Tp < class T > void write(T x) {
		if(O > LIMIT) flush() ; if(x < 0) { quq[O ++] = 0x2d ; x = -x ; }
		do { ST[++ sz] = x % 0xa ^ 0x30 ; x /= 0xa ; } while(x) ;
		while(sz) quq[O ++] = ST[sz --] ; return ;
	}
	FILEOUT & operator << (int x) { return write(x) , * this ; }
	FILEOUT & operator << (signed x) { return write(x) , * this ; }
	FILEOUT & operator << (unsigned x) { return write(x) , * this ; }
} out ;

int n , k ;
const int maxn = 1e5 + 10 ;
int a[maxn] ;
int dp[22][maxn] ;
int times = 1 ;
int L = 1 , R = 0 , val = 0 ;
int cnt[maxn] ;
void calc(int l , int r) {
	while(L > l) val += cnt[a[-- L]] ++ ;
	while(R < r) val += cnt[a[++ R]] ++ ;
	while(L < l) val -= -- cnt[a[L ++]] ;
	while(R > r) val -= -- cnt[a[R --]] ;
}
void solve(int l , int r , int a , int b) {
	if(l > r) return ;
	int p = 0 , mid = l + r >> 1 ;
	rep(i , a , min(b , mid)) {
		calc(i , mid) ;
		if(val + dp[times - 1][i - 1] < dp[times][mid]) {
			dp[times][mid] = val + dp[times - 1][i - 1] ;
			p = i ;
		}
	}
	solve(l , mid - 1 , a , p) ;
	solve(mid + 1 , r , p , b) ;
}
signed main() {
#ifdef _WIN64
	freopen("testdata.in" , "r" , stdin) ;
#else
	ios_base :: sync_with_stdio(false) ;
	cin.tie(nullptr) , cout.tie(nullptr) ;
#endif
// code begin.
	in >> n >> k ;
	rep(i , 1 , n) 
		in >> a[i] ;
	memset(dp , 0x3f , sizeof(dp)) ;
	dp[0][0] = 0 ;
	for( ; times <= k ; times ++)
		solve(1 , n , 1 , n) ;
	out << dp[k][n] << '\n' ;
	return out.flush() , 0 ;
// code end.
}