CF1312E Array Shrinking（区间dp）

Array Shrinking

思路：可以看出是区间dp的问题，n = 500，刚好可以是O（n^3）。难点在于相邻数字合并的维护，这里我们可以定义一个二维的数组a[l][r]，表示L到R区间合并后的数字是a[l][r]，比如 3 3 3 a[1][2] = 4, a[2][3] = 4，然后dp[l][r]表示L到R的最短数列长度，容易想到当dp[l][mid-1]和dp[mid][r]等于1且a[l][mid-1] == a[mid][r]时，区间可以合并。可以推出状态转移方程：

①(dp[l][mid-1] == 1 && dp[mid][r] == 1 && a[l][mid-1] == a[mid][r]) dp[l][r] = 1, a[l][r] = a[mid][r] + 1;

②dp[l][r] = min(dp[l][mid_1 - 1] + dp[mid_1][r], dp[l][mid_2 - 1] + dp[mid_2][r]...);

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <set>
#include <vector>
#include <queue>
#include <cstring>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second
 
const int N = 2e5 + 10;
const int INF = 1e9;
const int MAX_DIS = 2e6;

int _case = 0;

void solve(){
    int n;
    cin >> n;
    vector<vector<int > > num(n + 1, vector<int >(n + 1));
    for(int i = 1; i <= n; ++i) cin >> num[i][i];
    vector<vector<int > > dp(n + 1, vector<int >(n + 1, INF));
    for(int i = 1; i <= n; ++i) dp[i][i] = 1;
    for(int len = 2; len <= n; ++len){
        for(int l = 1; l <= n - len + 1; ++l){
            int r = l + len - 1;
            for(int mid = l + 1; mid <= n; ++mid){
                if(dp[l][mid - 1] == 1 && dp[mid][r] == 1 && 
                   num[l][mid - 1] == num[mid][r]){

                    dp[l][r] = 1;
                    num[l][r] = num[mid][r] + 1;
                }else{
                    dp[l][r] =
                    min(dp[l][r], dp[l][mid - 1] + dp[mid][r]);
                }
            }
        }
    }
    //cout << "ans  =";
    cout << dp[1][n] << endl;
}

int main(){
    
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    
    return 0;
}