HDU 5074 Hatsune Miku

Hatsune Miku

Time Limit: 1000ms
Memory Limit: 262144KB
This problem will be judged on HDU. Original ID: 5074
64-bit integer IO format: %I64d      Java class name: Main
 
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, . . . , an, is simply the sum of score(ai, ai+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, . . . , an (-1 ≤ ai ≤ m, ai ≠ 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

 
解题:动态规划乱搞
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 110;
 4 int dp[maxn][maxn],score[maxn][maxn],song[maxn];
 5 int main() {
 6     int n,m,kase;
 7     scanf("%d",&kase);
 8     while(kase--) {
 9         scanf("%d%d",&n,&m);
10         for(int i = 1; i <= m; ++i)
11             for(int j = 1; j <= m; ++j)
12                 scanf("%d",score[i] + j);
13         for(int i = 1; i <= n; ++i) scanf("%d",song + i);
14         memset(dp,-1,sizeof dp);
15         dp[0][0] = 0;
16         for(int i = 1; i <= n; ++i) {
17             for(int j = 0; j <= m; ++j) {
18                 if(dp[i-1][j] == -1) continue;
19                 if(song[i] == -1) {
20                     for(int k = 1; k <= m; ++k)
21                         dp[i][k] = max(dp[i][k],dp[i-1][j] + score[j][k]);
22                 } else dp[i][song[i]] = max(dp[i][song[i]],dp[i-1][j] + score[j][song[i]]);
23             }
24         }
25         int ret = 0;
26         if(song[n] == -1) {
27             for(int i = 1; i <= m; ++i) ret = max(ret,dp[n][i]);
28         } else ret = dp[n][song[n]];
29         printf("%d\n",ret);
30     }
31     return 0;
32 }
View Code

 

posted @ 2015-10-19 20:09  狂徒归来  阅读(331)  评论(0编辑  收藏  举报