HDU 5074 Hatsune Miku（dp）

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

25 383 86 7715 93 3586 92 493 3 3 1 210 536 11 68 67 2982 30 62 23 6735 29 2 22 5869 67 93 56 1142 29 73 21 19-1 -1 5 -1 4 -1 -1 -1 4 -1

 

Sample Output

270625

 

题意：（引用别人的解释）

  题意为有m种音符，编号1到m，我们要用这m种音符来创造一首带有n个音符的曲子（当然，一种音符可以用多次）,假设有两个连续的音符 i ,j ，那么定义score(i,j)为这两个音符的得分，题目中预先给出所有的score(i,j)   1<=i,j<=m, 那么我们创造出的n个音符的曲子的得分为  score( note[i] , note[i+1] ) + score (note[i+1] ,note[i+2) +......score(note[n-1],note[n]) ， i从1开始。 note [i] 代表n个音符的曲子中第i个音符是第几种音符， 1<=note[i]<=m， 比如 note[i]=3，就表示n个音符中第i个位置用的是第3种音符（一共有m种）,预先给出 这n个音符，note[1] 到note[n] ，其中note[ i] 或者等于-1 ，或者 大于等于1小于等于m，对于后者，该位置的音符不能改变，对于前者，该位置可以换成任意的音符j(1<=j<=m)， 问 这n个音符所获得的最大得分是多少。

解释在代码中，这里我贴出来一个对的和一个有bug的代码：

对的代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8using namespace std;#define N 105int b[N][N],a[N];int n,m;int dp[N][N];   //dp[i][j]表示长度为i,以j结尾的最大价值void solve(){int i,j,k;memset(dp,0,sizeof(dp));for(i=2;i<=n;i++)   if(a[i]<0)   {       if(a[i-1]<0)         {             for(j=1;j<=m;j++)       for(k=1;k<=m;k++)     dp[i][j]=max(dp[i][j],dp[i-1][k]+b[k][j]);         }         else            {               for(k=1;k<=m;k++)                  dp[i][k]=max(dp[i][k],dp[i-1][a[i-1]]+b[a[i-1]][k]);            }   }   else   {           if(a[i-1]>0)       {                dp[i][a[i]]=dp[i-1][a[i-1]]+b[a[i-1]][a[i]];       }       else            {                for(k=1;k<=m;k++)                dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][k]+b[k][a[i]]);            }   }}int main(){int i,j,t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(b,0,sizeof(b));for(i=1;i<=m;i++)for(j=1;j<=m;j++) scanf("%d",&b[i][j]);for(i=1;i<=n;i++)scanf("%d",&a[i]);solve();int ans=0;for(i=1;i<=m;i++)ans=max(ans,dp[n][i]);printf("%d\n",ans);}    return 0;}

有bug代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8using namespace std;#define N 105int b[N][N],a[N];int n,m;int dp[N][N];void solve(){    int i,j,k;    memset(dp,0,sizeof(dp));    for(i=2;i<=n;i++)       if(a[i]<0)     //bug就在这里，没有对a[i-1]分类，就是a[i-1]为定值，虽然dp[i-1][x]为0，但是b[x][j]//很大,所以需要严格的分类       {             for(j=1;j<=m;j++)              for(k=1;k<=m;k++)                dp[i][j]=max(dp[i][j],dp[i-1][k]+b[k][j]);       }       else       {           if(a[i-1]>0)               {                dp[i][a[i]]=dp[i-1][a[i-1]]+b[a[i-1]][a[i]];               }               else            {                for(k=1;k<=m;k++)                dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][k]+b[k][a[i]]);            }       }}int main(){    int i,j,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        memset(b,0,sizeof(b));                for(i=1;i<=m;i++)            for(j=1;j<=m;j++)             scanf("%d",&b[i][j]);        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        solve();        int ans=0;        for(i=1;i<=m;i++)            ans=max(ans,dp[n][i]);        printf("%d\n",ans);    }    return 0;}