【Coderforces 833 B. The Bakery】 线段树+dp

Codeforces 833B
题意：把n个数分成k部分，使得每部分价值之和最大。每部分的价值为不同数的个数

$$\begin{gathered} \text{我们首先知道这题很容易用}dp\left[i\right]\left[k\right]\text{想到是以 }i\text{结尾 前面已经有}k\text{部分的方案数} \\ \text{那么}dp\left[i\right]\left[k\right]=\max \left(dp\left[i\right]\left[k-1\right],dp\left[j\right]\left[k-1\right]+num\left[j+1\right]\left[i\right]\right) \\ \text{这样直接跑是}O\left(n^{2}m\right)\text{的 会}TLE\text{如果我们把取}\max x\text{用一个数组记录下来} \\ \text{那么每次决策就是}O\left(1\right)\text{的转移 但是这不是类似层次的东西} \\ \text{所以我们可以建一颗线段树} \\ \text{叶子节点的值是 }dp\left[j\right]\left[k-1\right]+num\left[j+1\right]\left[i\right] \\ \text{可是怎么更新}num\text{呢？对于这种不同数的问题} \\ \text{我们很容易想到}pre\text{数组 我们不考虑单个的}num \\ \text{我们考虑全局}num\text{每个数字对一段区间的贡献就是 }pre\left[i\right],i-\text{1 这段会加}1 \\ \text{于是更新部分就结束了 就可以开始}query\text{了 因为 }k-1<=j<i\text{所以}j\text{范围也有了} \\ i\text{每次要从}k\text{开始 不然前面不能凑齐}k-\text{1个} \end{gathered}$$

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 35025;
int maxx[MAX_N<<2],col[MAX_N<<2],pre[MAX_N],last[MAX_N],arr[MAX_N],dp[MAX_N][55];
set<int > st;
void up(int rt)
{
    maxx[rt] = max(maxx[rt<<1],maxx[rt<<1|1]);
}
void down(int rt,int l,int r)
{
    if(col[rt])
    {
        col[rt<<1] += col[rt];
        col[rt<<1|1] += col[rt];
        maxx[rt<<1] += col[rt];
        maxx[rt<<1|1] += col[rt];
        col[rt] = 0;
    }
}
void build(int rt,int l,int r,int v)
{
    col[rt] = maxx[rt] = 0;
    if(l==r)
    {
        maxx[rt] = dp[l][v];
        return;
    }
    build(rt<<1,l,mid,v);
    build(rt<<1|1,mid+1,r,v);
    up(rt);
}
void update(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        col[rt]++;
        maxx[rt]++;
        return;
    }
    down(rt,l,r);
    if(x<=mid) update(rt<<1,l,mid,x,y);
    if(mid<y) update(rt<<1|1,mid+1,r,x,y);
    up(rt);
}
int query(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        return maxx[rt];
    }
    down(rt,l,r);
    if(y<=mid) return query(rt<<1,l,mid,x,y);
    else if(x>mid) return query(rt<<1|1,mid+1,r,x,y);
    else return max(query(rt<<1,l,mid,x,y),query(rt<<1|1,mid+1,r,x,y));
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),pre[i] = last[arr[i]],last[arr[i]] = i,st.insert(arr[i]),dp[i][1] = st.size();
    for(int k = 2;k<=m;++k)
    {
        build(1,1,n,k-1);
        for(int i = k;i<=n;++i)
        {
            update(1,1,n,pre[i],i-1);
            dp[i][k] = max(dp[i][k-1],query(1,1,n,k-1,i-1));
        }
    }
    printf("%d\n",dp[n][m]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}