HDU 5074 Hatsune Miku

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5074

Surface:

Hatsune Miku Time limit:2000/1000 MS (java/others) Memory limit:262144/262144 K (java/others)
Total submission (s): 810 Accepted Submission (s): 577


Problem Descriptionhatsune Miku is a popular virtual singer. It is very popular in both Japan and China. Basically it is a computer software this 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 is only m different notes provided in the package. And you want the song with n notes.


Also, know that there are a system to evaluate the beautifulness of a song. For each of the consecutive notes A and B, if B comes after a, then the beautifulness for these the notes are evaluated as SCO Re (A, b).

So the total beautifulness for a song consisting of notes A1, A2, ..., an, was simply the sum of score (AI, ai+1) for 1≤ I≤n-1.

Now, your find that at some positions, the notes has to be some specific ones, but at other positions can decide what Notes to use. You want to maximize your song ' s beautifulness. What's the maximum beautifulness you can achieve?
Inputthe first line contains an integer T (t≤10), denoting the number of the the test cases.

For each test case, the first line contains the 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 Y OU cannot change, while negative integers is what you can replace with arbitrary notes. The notes is named from 1 to M.
Outputfor each test case, output the answer in one line.
Sample Input

25 383 86 7715 93 3586 92 493 3 3 1 210 536 11 68 67 2982 30 62 23 6735 29 2 22 5869 67 93 56 1142 29 73 21 19-1-1 5-1 4 -1-1-1 4-1

Sample Output

270625

Source2014 Asia Anshan Regional Contest

Solving:

DP[I][J] represents the maximum value of J selected at position I. The outermost loop I represents the current position, if I is not fixed, a double-layer cycle, respectively cyclic Dp[i+1][j]=max (dp[i+1][j],dp[i][k]+val[k][j]); If the current point is fixed, the single-layer loop K,dp[i+1][j]=max ( DP[I+1][J],DP[I][K]+VAL[K][J]). Also consider whether the subsequent points have been fixed.

Code:

#include <iostream> #include <cstring> #include <cstdio> #include <cstdlib>using namespace std; int val[55][55],dp[105][55],pos[105],p,ans;int t,n,m;int max (int a,int b) {return  a>b?a:b;} int main () {    int t;scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&m); Memset (Val,0,sizeof (Val)); memset (Dp,0,sizeof (DP)); for (int. i=1;i<=m;i++) {for (int j=1;j<=m;j++) scanf ("%d", &val[i][j]);} for (int i=1;i<=n;i++) {scanf ("%d", &pos[i]);} for (int i=2;i<=n;i++) {p=pos[i];if (pos[i]>0) {if (pos[i-1]>0)    dp[i][p]=dp[i-1][pos[i-1]]+val[pos[i-1] ][p];elsefor (int j=1;j<=m;j++) Dp[i][p]=max (Dp[i][p],dp[i-1][j]+val[j][p]);} Else{if (pos[i-1]>0) for    (int j=1;j<=m;j++) dp[i][j]=dp[i-1][pos[i-1]]+val[pos[i-1]][j];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]+val[k][j]);}} ans=0;for (int i=1;i<=m;i++) Ans=max (Ans,dp[n][i]);p rintf ("%d\n", ans);} return 0;}

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HDU 5074 Hatsune Miku