CodeForces - 1312E Array Shrinking【区间DP】

The meaning of problems

Given the array $a_1, a_2, ...,a_n$You can do the following to the array any number of times :

• Selecting a pair of adjacent and equal in number, $a_i = a_{i+1}$
• use $a_i+1$ instead of the logarithmic

After several arrays now seek operation, the shortest possible length.

$1 \ k \ le500, 1 \ The a_i \ le1000$

5
4 3 2 2 3

2

7
3 3 4 4 4 3 3

2

3
1 3 5

3

1
1000

1

answer

• status: $dp[i][j]$ represents the [i, j] array may reduce the range of the shortest length
• An array of additional $w[i][j]$ indicatesif [i, j] can be shortened to a length in the range of a value of 1, if dp [i] [j] is not one that w [i] [j] does not make sense. The determination for merging two array section
• State transition: $dp[i][j] = min(dp[i][k] + dp[k+1][j]), k∈[i,j]$
• If there is k such that $dp[i][k]==1, \ dp[k+1][j]==1, \ w[i][k]==w[k+1][j]$ Well $dp[i][j]=2$

#include<cstdio>
#include <iostream>
using namespace std;
#define ll long long
#define pr pair<int, int>
const int maxn=4000;

int n, m, ans;

int a[maxn];
int dp[maxn][maxn], val[maxn][maxn];

int main()
{
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> a[i];
        val[i][i] = a[i]; dp[i][i] = 1;
    }
    for(int i=n-1;i>0;i--)
    {
        for(int j=i+1;j<=n;j++)
        {
            dp[i][j] = 0x3f3f3f3f;
            for(int k=i;k<j;k++)
            {
                if(dp[i][k]+dp[k+1][j] < dp[i][j])
                {
                    dp[i][j] = dp[i][k] + dp[k+1][j];
                    if(val[i][k]==val[k+1][j] && dp[i][k] == dp[k+1][j] && dp[i][k] == 1)
                    {
                        val[i][j] = val[i][k] + 1;
                        dp[i][j] = 1;
                    }
                }
            }
        }
    }
    cout << dp[1][n] << endl;

    return 0;
}