Acwing 322. Elimination block (interval dp)

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Meaning of the question: give you a sequence of length n, you can choose a continuous block each time $[i,j]\; is$ eliminated, after elimination$i-1$ and$j+1$ connection, the score of this operation is$(j-i+1{)}^{2.}$ Ask how many points you can get.

Idea: It is easy to think of $f\left[i\right]\left[j\right]$ represents the interval$[i,j]\; is$ the maximum score, but if the$j+If\; 1\; is\; the$ same as the jth, we can’t just eliminate the$j$ without eliminating the$j+1$ piece. So we need to add one more dimension to represent the state, that is,$f\left[i\right]\left[j\right]\left[k\right]$ means to eliminate$[i,j]$ square and also eliminatejjAfter$j$ (not including$j$）$k$ andjjThe maximum value when$j$ blocks are the same block.

Then there are two types of state transitions:

1. From $j$ forward to find the last one and$j$ squares with different squares, record their position as$p$，则有$f[i][j][k]=max(f[i][j][k],f[i][p-1][0]+(j+k-p+1{)}^2)$
2. After finding the above p, we start from $i$ look backwards, find every q, there is$a[q]!=a[q+1]$&&$a[q]==a\left[j\right]$ , and satisfy$q+1<=p-1$，维护状态$f[i][j][k]=max(f[i][j][k],f[i][q][j+k-p+1]+f[q+1][p-1][0])$

Pay attention to the condition of recursive exit, and use recursive implementation.

Code:

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ls p<<1
#define rs p<<1|1
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ll long long
//#define int long long
#define pii pair<int,int>
#define ull unsigned long long
#define pdd pair<double,double>
#define lowbit(x) x&-x
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
char *fs,*ft,buf[1<<20];
#define gc() (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<20,stdin),fs==ft))?0:*fs++;
inline int read()
{
    
    
    int x=0,f=1;
    char ch=gc();
    while(ch<'0'||ch>'9')
    {
    
    
        if(ch=='-')
            f=-1;
        ch=gc();
    }
    while(ch>='0'&&ch<='9')
    {
    
    
        x=x*10+ch-'0';
        ch=gc();
    }
    return x*f;
}
using namespace std;
const int N=2e3+55;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1);

int a[N],f[222][222][222],cas=0;

int dfs(int x,int y,int z)
{
    
    
    if(f[x][y][z])
        return f[x][y][z];
    if(x==y)
        return f[x][y][z]=(1+z)*(1+z);
    if(x>y)
        return 0;
    int p=y;
    while(p>=x&&a[p]==a[y])
        p--;
    p++;
    f[x][y][z]=max(f[x][y][z],dfs(x,p-1,0)+(y+z-p+1)*(y+z-p+1));
    int q=x;
    while(q+1<=p-1)
    {
    
    
        if(a[q]==a[y]&&a[q+1]!=a[y])
            f[x][y][z]=max(f[x][y][z],dfs(x,q,z+y-p+1)+dfs(q+1,p-1,0));
        q++;
    }
    return f[x][y][z];
}
void solve()
{
    
    
    int n;
    cin>>n;
    memset(f,0,sizeof f);
    for(int i=1; i<=n; i++)
        cin>>a[i];
    cout<<"Case "<<++cas<<": "<<dfs(1,n,0)<<endl;
}
signed main()
{
    
    
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}