codeforces#426(div1) B - The Bakery (线段树 + dp)

B. The Bakery

 

Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which can bake several types of cakes, and opened the bakery.

Soon the expenses started to overcome the income, so Slastyona decided to study the sweets market. She learned it's profitable to pack cakes in boxes, and that the more distinct cake types a box contains (let's denote this number as the value of the box), the higher price it has.

She needs to change the production technology! The problem is that the oven chooses the cake types on its own and Slastyona can't affect it. However, she knows the types and order of n cakes the oven is going to bake today. Slastyona has to pack exactly k boxes with cakes today, and she has to put in each box several (at least one) cakes the oven produced one right after another (in other words, she has to put in a box a continuous segment of cakes).

Slastyona wants to maximize the total value of all boxes with cakes. Help her determine this maximum possible total value.

Input

The first line contains two integers n and k (1 ≤ n ≤ 35000, 1 ≤ k ≤ min(n, 50)) – the number of cakes and the number of boxes, respectively.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) – the types of cakes in the order the oven bakes them.

Output

Print the only integer – the maximum total value of all boxes with cakes.

Examples

input

4 1
1 2 2 1

output

2

input

7 2
1 3 3 1 4 4 4

output

5

input

8 3
7 7 8 7 7 8 1 7

output

6

Note

In the first example Slastyona has only one box. She has to put all cakes in it, so that there are two types of cakes in the box, so the value is equal to 2.

In the second example it is profitable to put the first two cakes in the first box, and all the rest in the second. There are two distinct types in the first box, and three in the second box then, so the total value is 5.

题意：

把 n 个数划分成 m 段，要求每组数不相等的数的数量最大之和。

思路：

dp方程 ： dp[i][j] = max( dp[k][j-1] + v(k, i) );( j<=k<i , k = j, j+1, +...+ i-1)

dp[i][j]表示第 i 个数分到第 j 段的最大值。

v(k, i) 表示k~i中不同数的个数，此处用hash记录每个数上一次出现的位置，从上一次出现的位置到当前位置的 dp[i][j-1] 值均可+1。

线段树优化：维护 dp[k~i][j-1]的最大值。
此时时间复杂度 O(n*m*log(n))。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 35010, maxk = 55, INF = 0x3f3f3f3f;

int a[maxn], dp[maxn][maxk], last[maxn], now[maxn];
int maxv[maxn<<2], lazy[maxn<<2];

push_down(int rt)
{
    if(lazy[rt])
    {
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        maxv[rt<<1]+=lazy[rt];
        maxv[rt<<1|1]+=lazy[rt];
        lazy[rt]=0;
    }
}

push_up(int rt)
{
    maxv[rt]=max(maxv[rt<<1], maxv[rt<<1|1]);
}

void update(int l, int r, int rt, int ul, int ur, int v)
{
    if(ul<=l && r<=ur)
    {
        maxv[rt]+=v;
        lazy[rt]+=v;
        return;
    }
    push_down(rt);
    int m=(l+r)/2;
    if(ul<=m) update(lson, ul, ur, v);
    if(m<ur) update(rson, ul, ur, v);
    push_up(rt);
}

int query(int l, int r, int rt, int ql, int qr)
{
    if(ql<=l && r<=qr)
        return maxv[rt];
    push_down(rt);
    int m=(l+r)/2, res=-INF;
    if(ql<=m) res=max(res, query(lson, ql, qr));
    if(qr>m) res=max(res, query(rson, ql, qr));
    push_up(rt);
    return res;
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; ++i)
        scanf("%d", &a[i]);
    for(int i=1; i<=n; ++i)
    {
        last[i]=now[a[i]];
        now[a[i]]=i;
    }
    for(int j=1; j<=m; ++j)
    {
        if(j!=1)
        {
            memset(maxv, 0, sizeof maxv);
            memset(lazy, 0, sizeof lazy);
            for(int i=j-1; i<=n; ++i)
            {
                update(0, n, 1, i, i, dp[i][j-1]);
            }
        }
        update(0, n, 1, max(j-1, last[j]), j-1, 1);
        dp[j][j]=j;
        for(int i=j+1; i<=n; ++i)
        {
            update(0, n, 1, max(j-1, last[i]), i-1, 1);
            dp[i][j]=query(0, n, 1, j-1, i-1);
        }
    }
    printf("%d\n", dp[n][m]);
    return 0;
}