1114D . Flood Fill

这道题还可以转化成最长回文串，实际的答案是 离散化后的数组最长回文串切成一半有几个数字，这些数字就是中间的数字需要对齐的对象，所以不用改变

#include <bits/stdc++.h>
using namespace std;

const int maxN = 5008;

int n;
int dp[maxN][maxN];
vector<int> a(1), b;
vector<int> ans;

void input() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        int x; cin >> x;
        if (x != a.back()) a.push_back(x);
    }
    n = a.size() - 1;

    b = a;
    reverse(b.begin() + 1, b.end());
}

void solve() {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (a[i] == b[j]) {dp[i][j] = dp[i-1][j-1] + 1;}
            else {dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}
        }
    }
}

void output() {
    cout << n - (dp[n][n] + 1)/2 << '\n';
}

int main() {
    
    input();
    solve();
    output();
    
    return 0;
}