2025/10/27
第一代图灵机
一样的套路，考虑每一个右端点对应的最左边可以到哪里，显然是最小的 \(j\) 使得 \(\max\limits_{j \le k \le i}pre_k = j - 1\)。考虑线段树维护一个区间内的最大的答案和最大的 \(pre_i\)，但是发现当 \(pushup\) 的时候不好判断右区间对整个区间的贡献，所以需要维护一个区间中右半边区间对答案的贡献。不过最后可以发现一个区间内最大的答案也没什么用，没办法只通过它算出一个区间内指定一个最小 \(pre\) 的答案，还是只能用 \(pushup\) 算出来
Code
#include <iostream>
#include <set>using namespace std;
using ll = long long;
using pll = pair<ll, ll>;const int kN = 2e5 + 1;int n, m, q, c[kN];
set<int> s[kN];
ll a[kN];
struct Tr {int mx;ll val;
} tr[kN * 4];ll Pushup(int mx, int x, int l, int r) {if (l == r)return a[l] - a[max(mx, tr[x].mx)];int m = (l + r) / 2;if (mx > tr[x * 2].mx)return max(a[m] - a[mx], Pushup(mx, x * 2 + 1, m + 1, r));return max(tr[x].val, Pushup(mx, x * 2, l, m));
}
void Update(int p, int k, int x = 1, int l = 1, int r = n) {if (l == r)return tr[x].mx = k, void();int m = (l + r) / 2;if (p <= m)Update(p, k, x * 2, l, m);elseUpdate(p, k, x * 2 + 1, m + 1, r);tr[x].mx = max(tr[x * 2].mx, tr[x * 2 + 1].mx), tr[x].val = Pushup(tr[x * 2].mx, x * 2 + 1, m + 1, r);
}
ll Query(int mx, int nl, int nr, int x = 1, int l = 1, int r = n) {if (nl <= l && r <= nr)return Pushup(mx, x, l, r);int m = (l + r) / 2;ll res = 0;if (nl <= m)res = Query(mx, nl, nr, x * 2, l, m);if (nr > m)res = max(res, Query(max(mx, tr[x * 2].mx), nl, nr, x * 2 + 1, m + 1, r));return res;
}int main() {
#ifndef ONLINE_JUDGEfreopen("in", "r", stdin);freopen("out", "w", stdout);
#endifcin.tie(0)->sync_with_stdio(0);cin >> n >> m >> q;for (int i = 1; i <= m; i++)s[i].insert(0);for (int i = 1; i <= n; i++)cin >> a[i], a[i] += a[i - 1];for (int i = 1; i <= n; i++) {cin >> c[i];Update(i, *s[c[i]].rbegin()), s[c[i]].insert(i);}for (int op, l, r; q--;) {cin >> op >> l >> r;if (op - 1) {auto it = s[c[l]].find(l);if (next(it) != s[c[l]].end())Update(*next(it), *prev(it));s[c[l]].erase(l), c[l] = r;it = s[c[l]].lower_bound(l);if (it != s[c[l]].end())Update(*it, l);Update(l, *prev(it)), s[c[l]].insert(l);} else {cout << Query(l - 1, l, r) << '\n';}}return 0;
}