HDOJ5074 Hatsune Miku(dp)

最新推荐文章于 2019-06-05 22:16:17 发布

GKHack

最新推荐文章于 2019-06-05 22:16:17 发布

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分类专栏： HDOJ 动态规划(dp)

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本文深入探讨了AI在音视频处理领域的应用，包括视频分割、语义识别、自动驾驶、AR、SLAM等前沿技术，以及语音识别、变声等关键技术。同时，介绍了AI在音视频处理过程中的优化策略和实际案例。

Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 989 Accepted Submission(s): 688

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

有m个音符，让你创作一首含有n个音符的歌曲，使得歌曲得分最高。

状态转移方程：dp[i][a[i]] = max(dp[i][a[i]], dp[i - 1][j] + score[j][a[i]]) dp[i][k] = max(dp[i][k], dp[i - 1][j] + score[j][k])

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 105;
int score[MAXN][MAXN], dp[MAXN][MAXN], a[MAXN], n, m;
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	while(t--) {
		memset(dp, -1, sizeof(dp));
		scanf("%d%d", &n, &m);
		for(int i = 1; i <= m; ++i)
			for(int j = 1; j <= m; ++j)
				scanf("%d", &score[i][j]);
		for(int i = 1; i <= n; ++i)
			scanf("%d", &a[i]);
		if(a[1] != -1) dp[1][a[1]] = 0;
		else 
			for(int i = 1; i <= m; ++i)
				dp[1][i] = 0;
		for(int i = 2; i <= n; ++i)
			for(int j = 1; j <= m; ++j)
				if(dp[i - 1][j] != -1) {
					if(a[i] != -1) dp[i][a[i]] = max(dp[i][a[i]], dp[i - 1][j] + score[j][a[i]]);
					else 
						for(int k = 1; k <= m; ++k)
							dp[i][k] = max(dp[i][k], dp[i - 1][j] + score[j][k]);
				}
		int ans = -1;
		for(int i = 1; i <= m; ++i)
			ans = max(ans, dp[n][i]);
		printf("%d\n", ans);
	}
	return 0;
}