Codeforces 833B

D. The Bakery

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 11 2 2 1

output

2

input

7 21 3 3 1 4 4 4

output

5

input

8 37 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个数分成k段，每段的价值为这一段数里不同数字的个数，问价值和最大为多少。

【思路】

显然需要用到动态规划。我们用dp[i][j]表示前j个数分成i段价值和的最大值。容易得到状态转移方程为

dp[i][j]=max{dp[i-1][x]+剩下数的贡献} {1<=x< j}

过程用线段树维护一下区间最大值即可。

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 35005;struct SegTree{    int left, right, val, lazy;}seg[maxn<<2];int dp[55][maxn], pos[maxn], pre[maxn], a[maxn];void pushup(int rt){    seg[rt].val = max(seg[rt<<1].val, seg[rt<<1|1].val);    return ;}void pushdown(int rt){    seg[rt<<1].val += seg[rt].lazy;    seg[rt<<1|1].val += seg[rt].lazy;    seg[rt<<1].lazy += seg[rt].lazy;    seg[rt<<1|1].lazy += seg[rt].lazy;    seg[rt].lazy = 0;}void build(int pos, int left, int right, int rt){    seg[rt].lazy = 0;    if(left == right)    {        seg[rt].val = dp[pos][left-1];        return ;    }    int mid = (left + right) >> 1;    build(pos, left, mid, rt<<1);    build(pos, mid+1, right, rt<<1|1);    pushup(rt);}void update(int l, int r, int left, int right, int rt){    if(l <= left && r >= right)    {        seg[rt].val++;        seg[rt].lazy++;        return ;    }    pushdown(rt);    int mid = (left + right) >> 1;    if(l <= mid)        update(l, r, left, mid, rt<<1);    if(r > mid)        update(l, r, mid+1, right, rt<<1|1);    pushup(rt);}int query(int l, int r, int left, int right, int rt){    if(l <= left && r >= right)    {        return seg[rt].val;    }    pushdown(rt);    int mid = (left + right) >> 1;    int ans = 0;    if(l <= mid)        ans = max(ans, query(l, r, left, mid, rt<<1));    if(r > mid)        ans = max(ans, query(l, r, mid+1, right, rt<<1|1));    pushup(rt);    return ans;}int main(){    int n, k;    while(scanf("%d%d", &n, &k) != EOF)    {        memset(pos, 0, sizeof(pos));        for(int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            pre[i] = pos[a[i]]+1;            pos[a[i]] = i;        }        for(int i = 1; i <= k; i++)        {            build(i-1, 1, n, 1);            for(int j = 1; j <= n; j++)            {                update(pre[j], j, 1, n, 1);                dp[i][j] = query(1, j, 1, n, 1);            }        }        printf("%d\n", dp[k][n]);    }    return 0;}