DescriptionDescription
定义一个区间的代价为
∑i(ci2)∑i(ci2)
cici为ii出现的次数。
因为两个函数都是凸的啊。。
所以就有决策单调性了。。
然后分治一下。
计算贡献的时候要注意不要计算到两层区间内。(因为复杂度会炸掉。。。)
我是每次都先计算区间[l1,l][l1,l]的。。
#include <bits/stdc++.h>
using namespace std;
const int N = 202020;
const int M = 25;
typedef long long ll;
inline char get(void) {
static char buf[100000], *S = buf, *T = buf;
if (S == T) {
T = (S = buf) + fread(buf, 1, 100000, stdin);
if (S == T) return EOF;
}
return *S++;
}
inline void read(int &x) {
static char c; x = 0;
for (c = get(); c < '0' || c > '9'; c = get());
for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}
int a[N], p[N], vis[N];
ll dp[N][M];
int n, k, clc;
ll val;
inline void Solve(int k, int l, int r, int l1, int r1) {
if (l > r) return;
int mid = (l + r) >> 1, pos; dp[mid][k] = 1ll << 59;
for (int i = l + 1; i <= mid; i++) val += p[a[i]]++;
for (int i = l1; i <= min(mid, r1); i++) {
if (dp[i - 1][k - 1] + val < dp[mid][k]) {
dp[mid][k] = dp[i - 1][k - 1] + val;
pos = i;
}
val -= --p[a[i]];
}
for (int i = l1; i <= min(mid, r1); i++) val += p[a[i]]++;
for (int i = l + 1; i <= mid; i++) val -= --p[a[i]];
Solve(k, l, mid - 1, l1, pos);
for (int i = l1; i < pos; i++) val -= --p[a[i]];
for (int i = l + 1; i <= mid + 1; i++) val += p[a[i]]++;
Solve(k, mid + 1, r, pos, r1);
for (int i = l + 1; i <= mid + 1; i++) val -= --p[a[i]];
for (int i = l1; i < pos; i++) val += p[a[i]]++;
}
int main(void) {
freopen("1.in", "r", stdin);
read(n); read(k);
for (int i = 1; i <= n; i++) read(a[i]);
clc = 1; val = 0;
for (int i = 1; i <= n; i++) {
val += p[a[i]]++;
dp[i][1] = val;
}
for (int i = 1; i <= n; i++) p[i] = 0;
val = 0; p[a[1]] = 1;
for (int i = 2; i <= k; i++) {
Solve(i, 1, n, 1, n);
}
cout << dp[n][k] << endl;
return 0;
}