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.

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.

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

解題思路：

這道題是DP，一個數做出貢獻僅在上一次到這次之間，線段樹維護區間最大值，查詢O(logn)

代碼：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define lll spc<<1
 5 #define rrr spc<<1|1
 6 using std::max;
 7 struct trnt{
 8     int lzt;
 9     int mxs;
10 }tr[10000000],str;
11 int dp[100000];
12 int lst[100000];
13 int nw[100000];
14 int n,K;
15 void pushup(int spc)
16 {
17     tr[spc].mxs=max(tr[lll].mxs,tr[rrr].mxs);
18     return ;
19 }
20 void add(int spc,int v)
21 {
22     tr[spc].lzt+=v;
23     tr[spc].mxs+=v;
24     return ;
25 }
26 void pushdown(int spc)
27 {
28     if(tr[spc].lzt)
29     {
30         add(lll,tr[spc].lzt);
31         add(rrr,tr[spc].lzt);
32         tr[spc].lzt=0;
33     }
34     return ;
35 }
36 void build(int l,int r,int spc)
37 {
38     tr[spc]=str;
39     if(l==r)
40     {
41         tr[spc].mxs=dp[l-1];
42         return ;
43     }
44     int mid=(l+r)>>1;
45     build(l,mid,lll);
46     build(mid+1,r,rrr);
47     pushup(spc);
48     return ;
49 }
50 void update(int l,int r,int ll,int rr,int spc,int v)
51 {
52     if(l>rr||ll>r)
53         return ;
54     if(ll<=l&&r<=rr)
55     {
56         add(spc,v);
57         return ;
58     }
59     pushdown(spc);
60     int mid=(l+r)>>1;
61     update(l,mid,ll,rr,lll,v);
62     update(mid+1,r,ll,rr,rrr,v);
63     pushup(spc);
64     return ;
65 }
66 int query(int l,int r,int ll,int rr,int spc)
67 {
68     if(l>rr||ll>r)
69         return -0x3f3f3f3f;
70     if(ll<=l&&r<=rr)
71         return tr[spc].mxs;
72     pushdown(spc);
73     int mid=(l+r)>>1;
74     return max(query(l,mid,ll,rr,lll),query(mid+1,r,ll,rr,rrr));
75 }
76 int main()
77 {
78     scanf("%d%d",&n,&K);
79     for(int i=1;i<=n;i++)
80     {
81         int tmp;
82         scanf("%d",&tmp);
83         lst[i]=nw[tmp];
84         nw[tmp]=i;
85     }
86     for(int i=1;i<=K;i++)
87     {
88         build(1,n,1);
89         for(int j=0;j<=n;j++)
90             dp[j]=0;
91         for(int j=i;j<=n;j++)
92         {
93             update(1,n,lst[j]+1,j,1,1);
94             dp[j]=query(1,n,i,j,1);
95         }
96     }
97     printf("%d\n",dp[n]);
98     return 0;
99 }

CF-833B The Bakery（線段樹優化Dp）