BZOJ3747 POI2015 Kinoman

Description

A total of M films, numbered 1~m, the good-looking value of Part I movies is WI w_i. In n days (numbered from 1~n) a movie is shown every day, and the first day of the show is part fi f_i.

You can choose L,r (1<=l<=r<=n) and watch all the movies in l,l+1,..., R days. If you watch more than once in the same movie, you'll be bored and you won't be able to get a good look at the movie. So the sum of the good looking values of the movie you want to maximize and watch only once.

Input Line 1: Two integer n,m (1<=m<=n<=1000000). Line 2: contains n integers f1,f2,..., fn f_1,f_2,\ldots,f_n (1≤fi≤m) (1\le f_i\le m). Line 3: contains m integers w1,w2,..., wm w_1,w_2,\ldots,w_m (1≤wj≤106) (1\le w_j \le10^6). Data range:
There are 20% data, n≤8000, 70% data, n≤100000, 100% data, n≤1000000.

Output

Output the maximum value of the sum of the good-looking values of the movie viewed and viewed only once .

Sample Input

9 4
2 3 1 1 4 1 2 4 1
5 3 6 6

Sample Output

15

Input Details

Watch 2nd, 3, ... the movie that was shown in \ldots,7 days, with the number of movies seen and seen only once is 2,3,4.

Solution:
The focus is on the use of the following sentence: "If you watch the same movie more than once, you will not be able to get the beautiful value of the film." ”

That is, for a selected interval, the weight of the element cannot be collected if there is a repetition of the element. Generally encountered "The problem of element not repeating within interval" We only need to judge the relationship between the current element and the next identical element . If the current element is in POS and the next same element is in Nxt[pos], then for all L<pos L =nxt[pos], the interval cannot get the Val value. If Pos<=r<nxt[pos] POS, the interval can be given a Val value.

That is, for the interval [pos,nxt[pos]−1] [pos,nxt[pos]-1] This interval prefix and will add a Val value. If we define a data structure that maintains a prefix that begins with the enumeration's L l as the end of the current position , the above operation is to update the interval. Obviously, those elements that are already smaller than l continue to go backwards: [pos,nxt[pos]−1]→[nxt[pos],nxt[nxt[pos]]−1] [Pos,nxt[pos]-1]\to[nxt[pos],nxt[nxt[pos]] -1] So the subject uses a line tree to maintain this relationship, split the time Complexity O (NLOGN)