cf 833B 线段树优化dp

写出转移方程即可。
贡献值计算有通用方法：记录每一个颜色上一个位置，这个颜色只在上个位置到当前位置有贡献，为1，这个可以通过线段树做到。同样，dp值也可以用线段树查询，所以我们把他们同时丢到一颗线段树里面就行了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
int read()
{
	char ch = getchar(); int x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 3e4 + 5100;
int pre[N];
int pos[N];
int n, k;
int a[N];
ll maxx[N<<2];
ll lazy[N<<2];
ll dp[N][55];
 
void pushup(int root)
{
	maxx[root] = max(maxx[lrt], maxx[rrt]);
}
void build(int l, int r, int root,int dep)
{
	lazy[root] = 0;
	if (l == r)
	{
		maxx[root] = dp[l][dep]; return;
	}
	int mid=(l + r) >> 1;
	build(lson,dep); build(rson,dep);
	pushup(root);
}
void pushdown(int root)
{
	if (lazy[root])
	{
		maxx[lrt] += lazy[root]; maxx[rrt] += lazy[root];
		lazy[lrt] += lazy[root];
		lazy[rrt] += lazy[root];
		lazy[root] = 0;
	}
}
void update(int l, int r, int root, int lf, int rt, ll val)
{
	if (lf <= l && r <= rt)
	{
		maxx[root] += val;
		lazy[root] += val;
		return;
	}
	pushdown(root);
	int mid = (l + r) >> 1;
	if (lf <= mid)update(lson, lf, rt, val);
	if (rt > mid)update(rson, lf, rt, val);
	pushup(root);
}
ll querry(int l, int r, int root, int lf, int rt)
{
	if (lf <= l && r <= rt)
	{
		return maxx[root];
	}
	pushdown(root);
	int mid = (l + r) >> 1;
	ll ans = 0;
	if (lf <= mid)ans = max(ans,querry(lson, lf, rt));
	if (rt > mid)ans = max(ans,querry(rson, lf, rt));
	return ans;
}
int main()
{
	n = read(); k = read();
	upd(i, 0, n)pos[i] = 0;
	upd(i, 0, n)dp[i][0] = 0;
	upd(i, 1, n)
	{
		a[i] = read(); pre[i] = pos[a[i]]; pos[a[i]] = i;
	}
	upd(j,1,k)
	{
		build(0, n, 1,j-1);
		upd(i, j, n)
		{
			update(0, n, 1, pre[i], i - 1, 1ll);
			dp[i][j] = querry(0, n,1, 0, i - 1);
		}
	}
	cout << dp[n][k] << endl;
	return 0;
}