CoderForces999F-Cards and Joy

F. Cards and Joy

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $n$ players sitting at the card table. Each player has a favorite number. The favorite number of the $j$-th player is $f_j$.

There are $k\cdot n$ cards on the table. Each card contains a single integer: the $i$-th card contains number $c_i$. Also, you are given a sequence $h_1,h_2,\dots ,h_k$. Its meaning will be explained below.

The players have to distribute all the cards in such a way that each of them will hold exactly $k$ cards. After all the cards are distributed, each player counts the number of cards he has that contains his favorite number. The joy level of a player equals $h_t$ if the player holds $t$ cards containing his favorite number. If a player gets no cards with his favorite number (i.e., $t=0$), his joy level is $0$.

Print the maximum possible total joy levels of the players after the cards are distributed. Note that the sequence $h_1,\dots ,h_k$ is the same for all the players.

Input

The first line of input contains two integers $n$ and $k$ ($1\le n\le 500,1\le k\le 10$) — the number of players and the number of cards each player will get.

The second line contains $k\cdot n$ integers $c_1,c_2,\dots ,c_{k\cdot n}$ ($1\le c_i\le {10}^5$) — the numbers written on the cards.

The third line contains $n$ integers $f_1,f_2,\dots ,f_n$ ($1\le f_j\le {10}^5$) — the favorite numbers of the players.

The fourth line contains $k$ integers $h_1,h_2,\dots ,h_k$ ($1\le h_t\le {10}^5$), where $h_t$ is the joy level of a player if he gets exactly $t$ cards with his favorite number written on them. It is guaranteed that the condition $h_{t-1}<h_t$ holds for each $t\in \left[\mathrm{2..}k\right]$.

Output

Print one integer — the maximum possible total joy levels of the players among all possible card distributions.

Examples

input

Copy

4 3
1 3 2 8 5 5 8 2 2 8 5 2
1 2 2 5
2 6 7

output

Copy

21

input

Copy

3 3
9 9 9 9 9 9 9 9 9
1 2 3
1 2 3

output

Copy

0

Note

In the first example, one possible optimal card distribution is the following:

• Player $1$ gets cards with numbers $[1,3,8]$;
• Player $2$ gets cards with numbers $[2,2,8]$;
• Player $3$ gets cards with numbers $[2,2,8]$;
• Player $4$ gets cards with numbers $[5,5,5]$.

Thus, the answer is $2+6+6+7=21$.

In the second example, no player can get a card with his favorite number. Thus, the answer is $0$.

题意：n∗kn∗k张卡片分给n个人，每人k张。第二行输入n∗kn∗k张卡片上面写的数字，第三行输入nn个人喜欢的数字，第四行输入k个数字，h[i]h[i]表示拿到ii张自己喜欢的卡片可以获得的快乐值。问所有人快乐值之和最大为多少？

题解：DP,dp[i][j],表示i个相同的数字给j个人(这j个人都喜欢这相同的数字)；cnt[i]表示数字i的数量，num[j]表示喜欢j 的人数；

状态转换方程为：dp[i][j]=max(dp[i][j],dp[i-u][j-1]+w[u]);(1=<u<=k)

AC代码为：

#include<bits/stdc++.h>
using namespace std;


const int maxn=1e5+10;
int n,k,a[maxn],f[5005],w[15];
int cnt[maxn],num[maxn],dp[5005][521];


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    memset(cnt,0,sizeof cnt);
    memset(num,0,sizeof num);
    cin>>n>>k;
    for(int i=1;i<=n*k;i++) cin>>a[i],cnt[a[i]]++;
    for(int i=1;i<=n;i++) cin>>f[i],num[f[i]]++;
    for(int i=1;i<=k;i++) cin>>w[i];
    
    for(int i=1;i<=n*k;i++)
    {
        dp[i][1]=w[min(i,k)];
        for(int j=2;j<=n;j++)
        {
            for(int u=1;u<=min(i,k);u++) 
                dp[i][j]=max(dp[i][j],dp[i-u][j-1]+w[u]);
        }
    }
    long long ans=0;
    for(int i=1;i<maxn;i++) if(num[i]) ans+=dp[cnt[i]][num[i]];
    cout<<ans<<endl;
    
    return 0;
}