hdu 5074 Hatsune Miku（2014 鞍山现场赛）

最新推荐文章于 2019-06-05 22:16:17 发布

原创最新推荐文章于 2019-06-05 22:16:17 发布 · 738 阅读

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本文探讨了音视频编解码算法的优化方法，包括视频容器、FFmpeg音视频编解码、音视频合并封装等关键技术。介绍了硬件编解码的优势，并讨论了如何通过高效的编解码技术提高音视频的质量和传输效率。

Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 17 Accepted Submission(s): 14

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

-1可以随便转换为1~m的值，求最大值，简单DP

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=150;
int map[maxn][maxn];
int dp[maxn][maxn];
int a[1000];
int main()
{
    int n,t,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",&map[i][j]);
            }
        }
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(int i=0; i<maxn; i++)
        {
            for(int j=0; j<maxn; j++)
            {
                dp[i][j]=-99999999;
            }
        }
        int ans=0;
        if(a[1]>0)
            dp[1][a[1]]=0;
        else
        {
            for(int i=1; i<=m; i++)
            {
                dp[1][i]=0;
                // printf("%d ",dp[1][i]);
            }
        }
        for(int i=2; i<=n; i++)
        {
            if(a[i]>0)
            {
                for(int j=1; j<=m; j++)
                {
                    dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][j]+map[j][a[i]]);
                }
            }
            else
            {
                for(int j=1; j<=m; j++)
                {
                    for(int k=1; k<=m; k++)
                    {
                        dp[i][j]=max(dp[i][j],dp[i-1][k]+map[k][j]);
                    }
                }
            }
            //printf("%d\n",dp[i][a[i]]);
        }
        for(int i=1; i<=m; i++)
            ans=max(ans,dp[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}