hdu Hatsune Miku(DP)

Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 162    Accepted Submission(s): 129

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

 

普通的二维dp，还是特别不熟悉，这里说一个需要注意的地方：

　　在进行dp[i][j]时，如果dp[i-1][j]为-1，此时应该注意，计算dp[i][j]不应将其计算在内。（这里可能存在dp[i-1][j] + max(score[j][k]) 最大的情况，故需要一个if来约束一下！）

 

状态dp[i][k]表示第i个位置放编号为k的物品

 

Code：

 1 //============================================================================ 2 // Name        : 1.cpp 3 // Author      :  4 // Version     : 5 // Copyright   : Your copyright notice 6 // Description : Hello World in C++, Ansi-style 7 //============================================================================ 8  9 #include <iostream>10 #include <algorithm>11 #include <stdio.h>12 #include <string.h>13 #include <math.h>14 #include <stdlib.h>15 #include <queue>16 using namespace std;17 18 int score[110][110],f[110];19 int dp[110][110];20 int t,n,k;21 22 int main() {23     //freopen("in.txt","r",stdin);24 25     scanf("%d",&t);26     while(t--){27         scanf("%d %d",&n,&k);28         for(int i=1;i<=k;i++) for(int j=1;j<=k;j++) scanf("%d",&score[i][j]);29         for(int i=1;i<=n;i++) scanf("%d",&f[i]);30 31         memset(dp,-1,sizeof(dp));32         if(f[1]==-1)33             for(int i=1;i<=k;i++) dp[1][i]=0;34         else dp[1][f[1]]=0;35         for(int i=2;i<=n;i++){36             if(f[i]==-1){37                 for(int z=1;z<=k;z++){38                     int maxc = -1;39                     for(int j=1;j<=k;j++)40                         if(dp[i-1][j] + score[j][z] > maxc && dp[i-1][j]!=-1) maxc = dp[i-1][j] + score[j][z];41                     dp[i][z] = maxc;42                 }43             }44             else{45                 //为特定数字的时候，比较特殊！46                 int maxc = -1;47                 int z = f[i];48                 if(f[i-1]!=-1) {49                     maxc = 0;50                     for(int l=1;l<=k;l++){51                         if(dp[i-1][l]==-1) maxc = maxc + 1;52                         maxc += dp[i-1][l];53                     }54                     maxc += score[f[i-1]][z];55                 }56                 else57                     for(int j=1;j<=k;j++)58                         if(dp[i-1][j] +score[j][z] > maxc && dp[i-1][j]!=-1) maxc = dp[i-1][j] +score[j][z];59                 dp[i][z] = maxc;60                 for(int zz=1;zz<=k;zz++)61                     if(zz!=z) dp[i][zz] = -1;62             }63         }64 //        for(int i=1;i<=n;i++){65 //            for(int j=1;j<=k;j++)66 //                printf("%d ",dp[i][j]);67 //            printf("\n");68 //        }69         int maxc = -1;70         for(int i=1;i<=k;i++)71             if(dp[n][i] >maxc) maxc = dp[n][i];72         printf("%d\n",maxc);73     }74 75     //cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!76     return 0;77 }