HDU 5074-Hatsune Miku（亚洲区域赛鞍山站 DP）

最新推荐文章于 2019-06-05 22:16:17 发布

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最新推荐文章于 2019-06-05 22:16:17 发布

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本文链接：https://blog.youkuaiyun.com/haoliang94/article/details/40618173

本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在不同场景下的实现方法，包括自动驾驶、AR增强现实等领域的应用实例。

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Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 29 Accepted Submission(s): 25

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

设状态dp[i][k]表示第i个位置放编号为k的物品，则dp[i][k]=max(dp[i][k],dp[i-1][j]+buff[j][k])

#include <cstdio>  
#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <cctype>  
#include <cmath>  
#include <cstdlib>  
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <list>  
#define ll long long  
using namespace std;  



int main()
{
    int i,j,k,l,m,n,t,ans;
    int dp[110][55];
    int a[110],buff[55][55];
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%d%d",&n,&m);
        memset(dp,-1,sizeof(dp));
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%d",&buff[i][j]);
            }
        }
        for(i=1;i<=n;i++)    scanf("%d",&a[i]);
        if(a[1]!=-1)    dp[1][a[1]]=0;
        else
        {
            for(i=1;i<=m;i++)
            {
                dp[1][i]=0;
            }
        }
        for(i=2;i<=n;i++)
        {
            for(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]+buff[j][a[i]]);
                    }
                    else
                    {
                        for(k=1;k<=m;k++)
                        {
                            dp[i][k]=max(dp[i][k],dp[i-1][j]+buff[j][k]);
                        }
                    }
                }
            }
        }
        for(i=1;i<=m;i++)
        {
            if(dp[n][i]>ans)    ans=dp[n][i];
        }
        printf("%d\n",ans);
        
    }    
}