codeforces 834 D. The Bakery

codeforces 834 D. The Bakery(dp + 线段树优化)

题意：

给一个长度为n的序列分成k段,每段的值为这一段不同数字的个数,最大化划分k端的值

$n <= 35000 \(
\)k <= min(n,50)$

思路:

由于k比较小，直接dp就好了

\(dp[i][j]\)选了k段到j的最大值

\(dp[i][j] = max(dp[i-1][k]+diff(k+1,j)) (0 <= k < j)\)

然后用线段树优化一下, 一个数的贡献是上一个相同数字+1的位置到当前位置

#include<bits/stdc++.h>
#define LL long long
#define P pair<int,int>
#define ls(i) seg[i].lc
#define rs(i) seg[i].rc
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ls rt<<1
#define rs (rt<<1|1)
using namespace std;
const int N = 4e4 + 10;
int read(){
    int x = 0;
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
    return x;
}
int n,k;
int a[N],pre[N],last[N];
int dp[55][N];
int col[N<<2],mx[N<<2];
void pushup(int rt){
    mx[rt] = max(mx[ls],mx[rs]);
}
void pushdown(int rt){
    if(col[rt]){
        col[ls] += col[rt],col[rs] += col[rt];
        mx[rs] += col[rt],mx[ls] += col[rt];
        col[rt] = 0;
    }
}
void update(int L,int R,int v,int l,int r,int rt){
    if(L <= l &&  R >= r){
        mx[rt] += v;
        col[rt] += v;
        return ;
    }
    pushdown(rt);
    int m = l + r>>1;
    if(L <= m) update(L,R,v,lson);
    if(R > m) update(L,R,v,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L <= l && R >= r) return mx[rt];
    pushdown(rt);
    int ans = 0;
    int m = l +r >>1;
    if(L <= m) ans = max(ans,query(L,R,lson));
    if(R > m) ans = max(ans,query(L,R,rson));
    return ans;
}
int main(){

    n = read(),k = read();
    for(int i = 2;i <= n + 1;i++){
        scanf("%d",a + i);
        pre[i] = last[a[i]];
        if(pre[i] == 0) pre[i] = 1;
        last[a[i]] = i;
    }
    for(int i = 1;i <= k;i++){
        for(int j = 2;j <= n + 1;j++){
            update(pre[j],j-1,1,1,n+1,1);
            dp[i][j] = query(pre[j],j-1,1,n+1,1);
        }
        for(int k = 1;k <= ((n+1)<<2);k++) mx[k] = col[k] = 0;
        for(int j = 1;j <= n + 1;j++){
            update(j,j,dp[i][j],1,n+1,1);
        }
    }
    int ans = 0;
    for(int j = 1;j <= n + 1;j++) ans = max(ans,dp[k][j]);
    cout<<ans<<endl;
    return 0;
}