CodeForces 834D The Bakery(线段树优化DP)

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：给出一个数组，将数组分割成k份，使每份中不同数字的个数加起来最大，输出这个最大值

题解：这道题很好列出状态转移方程，dp[i][j]=max{dp[k][j-1]+cnt[k+1][i]}

dp[i][j]意为1~i之间分割j次所产生的最大值。cnt[i][j]表示i-j之间不同的颜色个数。

看到max可以考虑线段树优化，也就是如果我们有办法在O(logn)的时间内计算出cnt[i][j]的值就可以了，考虑一种颜色能够产生贡献的范围，为他上一次出现的位置到他当前的位置，产生贡献为1，可以使用线段树区间加，给pre[i]~i都加上1，从1到n进行枚举，逐渐进行上面的操作，每次换下一层的时候重构线段树，这样计算cnt[i][j]的复杂度就变成了logn的

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lson root<<1
#define rson root<<1|1
using namespace std;

struct node
{
    int l,r,sum,lazy;
}tr[];
int dp[][],pre[],pos[];

void init()
{
    memset(tr,,sizeof(tr));
}

void push_up(int root)
{
    tr[root].sum=max(tr[lson].sum,tr[rson].sum);
}

void push_down(int root)
{
    tr[lson].sum+=tr[root].lazy;
    tr[lson].lazy+=tr[root].lazy;
    tr[rson].sum+=tr[root].lazy;
    tr[rson].lazy+=tr[root].lazy;
    tr[root].lazy=;
}

void build(int root,int l,int r,int now)
{
    if(l==r)
    {
        tr[root].l=l;
        tr[root].r=r;
        tr[root].sum=dp[now][l-];
        return ;
    }
    tr[root].l=l;
    tr[root].r=r;
    int mid=(l+r)>>;
    build(lson,l,mid,now);
    build(rson,mid+,r,now);
    push_up(root);
}

void update(int root,int l,int r,int val)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        tr[root].sum+=val;
        tr[root].lazy+=val;
        return ;
    }
    if(tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>;
    if(mid<l)
    {
        update(rson,l,r,val);
    }
    else
    {
        if(mid>=r)
        {
            update(lson,l,r,val);
        }
        else
        {
            update(lson,l,mid,val);
            update(rson,mid+,r,val);
        }
    }
    push_up(root);
}

int query(int root,int l,int r)
{
    if(tr[root].l==l&&tr[root].r==r)
    {
        return tr[root].sum;
    }
    if(tr[root].lazy)
    {
        push_down(root);
    }
    int mid=(tr[root].l+tr[root].r)>>;
    if(mid<l)
    {
        return query(rson,l,r);
    }
    else
    {
        if(mid>=r)
        {
            return query(lson,l,r);
        }
        else
        {
            return max(query(lson,l,mid),query(rson,mid+,r));
        }
    }
}

int main()
{
    int n,k,t;
    scanf("%d%d",&n,&k);
    for(int i=;i<=n;i++)
    {
        scanf("%d",&t);
        pre[i]=pos[t]+;
        pos[t]=i;
    }
    for(int i=;i<=k;i++)
    {
        init();
        build(,,n,i-);
        for(int j=;j<=n;j++)
        {
            update(,pre[j],j,);
            dp[i][j]=query(,,j);
        }
    }
    printf("%d\n",dp[k][n]);
}