CF 538 D. Flood Fill 递归 区间DP

link：https://codeforces.com/contest/1114/problem/D

题意：

　　给定一个数组，有不同的颜色，你可以从任意一个位置开始，改变颜色，相邻的是同一种颜色的位子的颜色也要跟着改变，问最少需要改变几次颜色。

思路：

　　我一开始想的是去掉相邻重复后，假设有k个，那么答案就是k-1个，然后可以使结果更优的就是不相邻相同的，还能使结果更优的是一层再套一层相同的，就比如1，2，3，2，1。现场不知道怎么数这个套了几层，经过高人代码点拨，就是记忆化递归。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

// priority_queue<int> q;//这是一个大根堆q
// priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
// #define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) // 用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);
// priority_queue<int ,vector<int>, greater<int> >que;

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  // 2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; // 18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; // 黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
/*-----------------------showtime----------------------*/
const int maxn = 5e3 + 9;
int a[maxn], b[maxn], dp[maxn][maxn];
int cal(int le, int ri) {
    if (dp[le][ri] >= 0) return dp[le][ri];
    if (le >= ri) return 0;
    if (b[le] == b[ri])
        return dp[le][ri] = cal(le + 1, ri - 1) + 1;
    else
        return dp[le][ri] = max(cal(le + 1, ri), cal(le, ri - 1));
}
int main() {
    int n;
    scanf("%d", &n);
    memset(dp, -1, sizeof(dp));
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

    int tot = 0;
    for (int i = 1, j; i <= n; i++) {
        for (j = i; a[i] == a[j] && j <= n; j++)
            ;
        b[++tot] = a[i];
        i = j - 1;
    }
    cout << tot - 1 - cal(1, tot) << endl;
    return 0;
}