hdoj 5074

Problem Description

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?

 

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases.

For each test case, the first line contains two integers n(1 ≤ n ≤ 100) and m(1 ≤ m ≤ 50) as mentioned above. Then m lines follow, each of them consisting of m space-separated integers, the j-th integer in the i-th line for score(i, j)( 0 ≤ score(i, j) ≤ 100). The next line contains n integers, a₁, a₂, . . . , a_n (-1 ≤ a_i ≤ m, a_i ≠ 0), where positive integers stand for the notes you cannot change, while negative integers are what you can replace with arbitrary notes. The notes are named from 1 to m.

 

Output

For each test case, output the answer in one line.

 

Sample Input

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

 

Sample Output

270 625

 

Source

2014 Asia AnShan Regional Contest

 

 分析：记dp[i][j]为第i位取第j位的最大值

#include "stdio.h"
#include "string.h"
#define MAX 110
int a[MAX];
int dp[MAX][MAX],g[MAX][MAX];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int t;
    int i,j,k,n,m;
    int ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1; i<=m; i++)
            for(j=1; j<=m; j++)
                scanf("%d",&g[i][j]);
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(i=2; i<=n; i++)
        {
            if(a[i]>0)
            {
                if(a[i-1]>0)
                     dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][a[i-1]]+g[a[i-1]][a[i]]);
                else
                {
                    for(j=1; j<=m; j++)
                        dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][j]+g[j][a[i]]);
                }
            }
            else
            {
                if(a[i-1]>0)
                {
                    for(j=1; j<=m; j++)
                        dp[i][j]=max(dp[i][j],dp[i-1][a[i-1]]+g[a[i-1]][j]);
                }
                else
                {
                    for(j=1; j<=m; j++)
                        for(k=1; k<=m; k++)
                            dp[i][k]=max(dp[i][k],dp[i-1][j]+g[j][k]);
                }
            }
        }
        ans=0;
        for(i=1; i<=m; i++)
            ans=max(ans,dp[n][i]);
        printf("%d
",ans);
    }
    return 0;
}