HDU 5074 Hatsune Miku （动态规划）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5074

题面：

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 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, . . . , an, is simply the sum of score(ai, ai+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, . . . , an (-1 ≤ ai ≤ m, ai ≠ 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


解题：

dp[i][j]代表的是在i位置选用j的最大值。最外面循环i代表当前位置，如果i不固定，双层循环，分别循环dp[i+1][j]=max(dp[i+1][j],dp[i][k]+val[k][j]);如果当前点已经固定了,那么单层循环k,dp[i+1][j]=max(dp[i+1][j],dp[i][k]+val[k][j])。同时需考虑后续点是否也已经固定了。

代码：

#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];
				else
					for(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
[i]);
		printf("%d\n",ans);
	}
	return 0;
}