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hdu5074 Hatsune Miku 2014鞍山現場賽E題水dp

編輯：C++入門知識

hdu5074 Hatsune Miku 2014鞍山現場賽E題水dp

Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 325 Accepted Submission(s): 243

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

Source 2014 Asia AnShan Regional Contest
題意：n個位置放樂符，總共m種樂符，給一個m*m的的矩陣表示(i,j)表示j在i後面一個位置產生的value，有的位置已經放了給定樂符，問所產生的最大value是多少。題解：水dp。dp[i][j]表示第i個位置放第j種樂符所產生的value,就是轉移的時候要注意下各種條件，約等於一個模擬題了。。。

/**
 * @author neko01
 */
//#pragma comment(linker, /STACK:102400000,102400000)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf(%d
,a)
typedef pair pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
int a[55][55];
int dp[105][55];
int b[105];
int main()
{
    int t;
    scanf(%d,&t);
    while(t--)
    {
        int n,m;
        scanf(%d%d,&n,&m);
        for(int i=1;i<=m;i++)
            for(int j=1;j<=m;j++)
                scanf(%d,&a[i][j]);
        b[0]=0;
        for(int i=1;i<=n;i++)
            scanf(%d,&b[i]);
        clr(dp);
        for(int i=2;i<=n;i++)
        {
            if(b[i]!=-1)
            {
                if(b[i-1]==-1)
                {
                    for(int j=1;j<=m;j++)
                    {
                        dp[i][b[i]]=max(dp[i-1][j]+a[j][b[i]],dp[i][b[i]]);
                    }
                }
                else
                    dp[i][b[i]]=dp[i-1][b[i-1]]+a[b[i-1]][b[i]];
            }
            else
            {
                if(b[i-1]==-1)
                {
                    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]+a[k][j]);
                }
                else
                    for(int j=1;j<=m;j++)
                        dp[i][j]=max(dp[i][j],dp[i-1][b[i-1]]+a[b[i-1]][j]);
            }
        }
        int ans=0;
        if(b[n]!=-1) ans=dp[n][b[n]];
        else
        {
            ans=dp[n][1];
            for(int i=2;i<=m;i++)
                ans=max(ans,dp[n][i]);
        }
        printf(%d
,ans);
    }
    return 0;
}