hdu 5074 Hatsune Miku 简单dp

题意：

解法：

代码：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)
#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)
#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])
#define mem(a,x)  memset(a,x,sizeof a)
#define ysk(x)  (1<<(x))
typedef long long ll;
typedef pair<int, int> pii;
const int INF =0x3f3f3f3f;
const int maxn=  100;
int n,m;
int sc[maxn+5][maxn+5],a[maxn+10];
int dp[maxn+5][maxn+5];
void solve()
{
mes(dp,-1,n+1);
for1(j,m)  if(a[1]==-1||a[1]==j)  dp[1][j]=0;

int ans=0;
for(int i=2;i<=n;i++)
{
for1(cur,m) if(a[i]==cur|| a[i]==-1  )
{
for1(k,m) if(~dp[i-1][k])
{
dp[i][cur]=max(dp[i][cur],dp[i-1][k]+sc[k][cur]);
}
if(i==n) ans=max(ans,dp[i][cur]);
}
}
cout<<ans<<endl;
}
int main()
{
std::ios::sync_with_stdio(false);
int T;cin>>T;
while(T--)
{
cin>>n>>m;
for1(i,m)  for1(j,m)  cin>>sc[i][j];
for1(i,n)  cin>>a[i];

solve();

}
return 0;
}