Hatsune Miku

Hatsune Miku is a popular virtual singer. It is very popular in both Japan and China. Basically it is a computer software that allows you to compose a song on your own using the vocal package.

Today you want to compose a song, which is just a sequence of notes. There are only m different notes provided in the package. And you want to make a song with n notes.


Also, you know that there is a system to evaluate the beautifulness of a song. For each two consecutive notes a and b, if b comes after a, then the beautifulness for these two notes is evaluated as score(a, b).

So the total beautifulness for a song consisting of notes a₁, a₂, . . . , a_n, is simply the sum of score(a_i, a_i+1) for 1 ≤ i ≤ n - 1.

Now, you find that at some positions, the notes have to be some specific ones, but at other positions you can decide what notes to use. You want to maximize your song’s beautifulness. What is the maximum beautifulness you can achieve?

2
5 3
83 86 77
15 93 35
86 92 49
3 3 3 1 2
10 5
36 11 68 67 29
82 30 62 23 67
35 29 2 22 58
69 67 93 56 11
42 29 73 21 19
-1 -1 5 -1 4 -1 -1 -1 4 -1

270
625

//1660K	1057B
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[107][57],scor[57][57],a[107];
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&scor[i][j]);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int i=2;i<=n;i++)
        {
            if(a[i]>0)
            {
                if(a[i-1]>0)//(a,b)
                {
                    dp[i][a[i]]=dp[i-1][a[i-1]]+scor[a[i-1]][a[i]];
                }
                else//(?,b)
                {
                    for(int j=1;j<=m;j++)
                    {
                        dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][j]+scor[j][a[i]]);
                    }
                }
            }
            else
            {
                for(int j=1;j<=m;j++)
                {
                    if(a[i-1]>0)//(a,?)
                    {
                        dp[i][j]=max(dp[i][j],dp[i-1][a[i-1]]+scor[a[i-1]][j]);
                    }
                    else//(?,?)
                    {
                        for(int k=1;k<=m;k++)
                        {
                            dp[i][j]=max(dp[i][j],dp[i-1][k]+scor[k][j]);
                        }
                    }
                }
            }
        }
        int ans=-1;
        for(int i=1;i<=m;i++)
            ans=max(ans,dp[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}