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;
}