HDU 5074 Hatsune Miku 暴力dp

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Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 677    Accepted Submission(s): 490

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 a ₁, a ₂, . . . , 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, a ₁, a ₂, . . . , 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

 

有m个音节，并给你每两个连续音节的分数，给你一个音节数列，如果数列中的值为-1，表示这个位置可以填任何音节，求此音节数列的最大分数。

dp[i][j]表示在第i个音节上是音符j的最大分数，共有四种情况：(a,b),(?,b),(a,?),(?,?)。分情况计算就可以了。

//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;
}