题解：P13339 [EGOI 2025] 礼品盒

注意到：若一支队伍的人数大于等于 \(3\)，则除第一个人与最后一个人以外，其他人必定会被跳过。这是因为只能跳过一段区间，如果一个中间位置的人没被跳过，那么要么她前面的所有人都没有被跳过，要么她后面的所有人都没有被跳过，这样的话必定有与她同队的人也获得奖品，这个队伍就获得至少两个奖品了。

对于区间 \(I_1=[l_1,r_1],I_2=[l_2,r_2]\)，我们定义：

\[\operatorname{merge}(I_1,I_2)=[\min\{l_1,l_2\},\max\{r_1,r_2\}] \]

即为删掉 \(I_1,I_2\) 所有元素所需的最短区间。

根据上面的结论，每个至少 \(3\) 个人的队伍都有一段必须要删除的区间，将它们 \(\operatorname{merge}\) 起来就得到了整体中必须要删除的区间 \([ql,qr]\)。

把 \([ql,qr]\) 的人都删除，剩余序列中每个队伍的人数都不超过 \(2\)。考虑对每个位置 \(l\)，求出最小的 \(r\)，使得删除 \([l,r]\) 之后每个队伍的人数都不超过 \(1\)，可以双指针维护。对于求出的每个 \([l,r]\)，实际需要删除的区间为 \(\operatorname{merge}([ql,qr],[l,r])\)，在所有这些区间中取最短的即为答案。

时间复杂度 \(O(n)\)。

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
    uniform_int_distribution<int> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

template<int mod>
inline unsigned int down(unsigned int x) {
	return x >= mod ? x - mod : x;
}

template<int mod>
struct Modint {
	unsigned int x;
	Modint() = default;
	Modint(unsigned int x) : x(x) {}
	friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
	friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
	friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
	friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
	friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
	friend Modint operator/(Modint a, Modint b) {return a * ~b;}
	friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
	friend Modint operator~(Modint a) {return a ^ (mod - 2);}
	friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
	friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
	friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
	friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
	friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
	friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
	friend Modint& operator++(Modint& a) {return a += 1;}
	friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
	friend Modint& operator--(Modint& a) {return a -= 1;}
	friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
	friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
	friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};

const int N = 5e5 + 5;

int t, n, a[N], cnt[N], rem[N];
vector<int> vec[N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> t >> n;
    rep(i, 0, n - 1) cin >> a[i];
    rep(i, 0, n - 1) vec[a[i]].push_back(i);
    int ql = n, qr = -1;
    rep(i, 0, t - 1) {
        int tot = vec[i].size();
        if(tot > 2) {
            int ul = vec[i][1], ur = vec[i][tot - 2];
            chkmin(ql, ul);
            chkmax(qr, ur);
        }
    }
    rep(i, ql, qr) a[i] = -1;
    rep(i, 0, n - 1) if(a[i] != -1) ++cnt[a[i]];
    int l = 0, r = -1, now = 0, ansl = -1, ansr = n;
    rep(i, 0, t - 1) if((rem[i] = cnt[i]) <= 1) ++now;
    while(l < n) {
        while(r + 1 < n && now < t) {
            ++r;
            if(a[r] != -1 && rem[a[r]]-- == 2) ++now;
        }
        if(now < t) break;
        int ul = min(l, ql), ur = max(r, qr);
        if(ur - ul + 1 < ansr - ansl + 1) {
            ansl = ul;
            ansr = ur;
        }
        if(a[l] != -1 && ++rem[a[l]] == 2) --now;
        ++l;
    }
    cout << ansl << " " << ansr << endl;
    return 0;
}