Codeforces 833B

B. The Bakery

time limit per test 2.5seconds

memory limit per test     256megabytes

 

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

Soon the expenses started to overcome the income, so Slastyonadecided to study the sweets market. She learned it's profitable to pack cakesin boxes, and that the moredistinct cake types a box contains (let's denote this number as the value of the box), the higherprice it has.

She needs to change the production technology! The problem isthat the oven chooses the cake types on its own and Slastyona can't affect it.However, she knows the types and order ofn 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 oneright after another (in other words, shehas to put in a box a continuous segment of cakes).

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

Input

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

The second line containsn integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 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 putall cakes in it, so that there are two types of cakes in the box, so the valueis equal to2.

In the second example it is profitable to put the first two cakes in the firstbox, and all the rest in the second. There are two distinct types in the firstbox, and three in the second box then, so the total value is5.

 

【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 <cstdio>#include <vector>#include <cstring>#include <algorithm>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define rush() int T;scanf("%d",&T);while(T--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long  ll;const int maxn = 35005;const ll mod = 1e9+7;const int INF = 0x3f3f3f;const double eps = 1e-9;int n,k;int dp[55][maxn];                 //表示到第j个数分成了i段的最大价值int tree[maxn<<2],lazy[maxn<<2];int a[maxn],pre[maxn],pos[maxn];void init(){    mst(tree,0),mst(pos,0);    mst(pre,0),mst(dp,0);}void pushup(int rt){    tree[rt]=max(tree[rt<<1|1],tree[rt<<1]);}void pushdown(int rt){    tree[rt<<1]+=lazy[rt];    tree[rt<<1|1]+=lazy[rt];    lazy[rt<<1]+=lazy[rt];    lazy[rt<<1|1]+=lazy[rt];    lazy[rt]=0;}void build(int pos,int l,int r,int rt){    lazy[rt]=0;    if(l==r)    {        tree[rt]=dp[pos][l-1];        return;    }    int m=(l+r)>>1;    build(pos,lson);    build(pos,rson);    pushup(rt);}void update(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        tree[rt]++;        lazy[rt]++;        return;    }    pushdown(rt);    int m=(l+r)>>1;    if(L<=m) update(L,R,lson);    if(R>m)  update(L,R,rson);    pushup(rt);}int query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return tree[rt];    }    pushdown(rt);    int m=(l+r)>>1;    int ans=0;    if(L<=m) ans=max(ans,query(L,R,lson));    if(R>m)  ans=max(ans,query(L,R,rson));    pushup(rt);    return ans;}int main(){    while(~scanf("%d%d",&n,&k))    {        init();        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);            pre[i]=pos[a[i]]+1;             //相同数字前一次出现的数字+1            pos[a[i]]=i;        }        for(int i=1;i<=k;i++)        {            build(i-1,1,n,1);               //在线段树中更新前一层dp的值            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;}