Codeforces（E. Array Shrinking）区间DP

Compare bare a range of DP
when the game did not launch D, this question did not even look. . .

Dp template directly set interval on the line
I ran twice, to run a D [] [], a run F [] []
D [I] [J] i ~ j represents the maximum can be incorporated into, if D [ i] [j] is not 0 also represents i ~ j can be combined into a number of
f [i] [j] represents the minimum length i ~ j

Code:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int a[1000],d[600][600],f[600][600],inf=99999999;
void init()
{
    for(int i=0;i<550;i++)
    {
        for(int j=0;j<=550;j++)
        {
            if(i==j)
            {f[i][j]=1;}
            else
            {f[i][j]=inf;}
        }
    }
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {cin>>a[i];d[i][i]=a[i];}
    for(int len=2;len<=n;len++)
    {
        for(int i=1;i<=n;i++)
        {
            int j=i+len-1;
            if(j>n){break;}
            for(int k=i;k<j;k++)
            {
                if(d[i][k]==d[k+1][j]&&d[i][k]!=0)
                {
                    d[i][j]=max(d[i][j],d[i][k]+1);
                }
            }
        }
    }
    if(d[1][n]!=0)
    {cout<<"1"<<endl;return 0;}
    init();
    for(int len=2;len<=n;len++)
    {
        for(int i=1;i<=n;i++)
        {
            int j=i+len-1;
            if(j>n){break;}
            for(int k=i;k<j;k++)
            {
                int l,r;
                if(d[i][k]!=0)
                {l=1;}
                else
                {l=k-i+1;}
                if(d[k+1][j]!=0)
                {r=1;}
                else
                {r=j-(k+1)+1;}
                if(len==2)
                {
                    if(d[i][k]==d[k+1][j])
                    {f[i][j]=1;}
                    else
                    {f[i][j]=2;}
                    continue;
                }
                f[i][j]=min(f[i][j],l+r);
                f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
                f[i][j]=min(f[i][j],l+f[k+1][j]);
                f[i][j]=min(f[i][j],f[i][k]+r);
            }
        }
    }
    cout<<f[1][n]<<endl;
    return 0;
}