题目链接:https://www.luogu.com.cn/problem/P4137
区间$mex$的查询的一道板子题,这里介绍一种非莫队的做法
考虑扫描线思想,先将所有询问存起来,从小到大$for$右端点$r$从$[1, n]$,同时用线段树维护出$pos[x]$表示数$x$上一次出现的位置,每次$r$移动时更新$pos[a[r]] = r$,则对于当前右端点$r$所有询问的左端点$l$的$mex$即为最小的$x$满足$pos[x] < l$,这个可以通过在线段树上二分实现
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int n, m;
int a[N];
int ans[N];
vector<array<int, 2>> que[N];
struct Node
{
int l, r;
int v;
} tr[N * 4];
void pushup(int u)
{
tr[u].v = min(tr[u << 1].v, tr[u << 1 | 1].v);
}
void build(int u, int l, int r)
{
tr[u] = {l, r};
if (l == r) return;
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int x, int c)
{
if (tr[u].l == x && tr[u].r == x) tr[u].v = c;
else
{
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, c);
else modify(u << 1 | 1, x, c);
pushup(u);
}
}
int query(int u, int c)
{
if (tr[u].v >= c) return -1;
if (tr[u].l == tr[u].r) return tr[u].r;
if (tr[u << 1].v < c) return query(u << 1, c);
return query(u << 1 | 1, c);
}
void solve()
{
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) cin >> a[i];
for (int i = 0; i < m; i ++ )
{
int l, r;
cin >> l >> r;
que[r].push_back({l, i});
}
build(1, 0, 2e5);
for (int r = 1; r <= n; r ++ )
{
modify(1, a[r], r);
for (auto [l, id] : que[r])
ans[id] = query(1, l);
}
for (int i = 0; i < m; i ++ )
cout << ans[i] << "\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T -- ) solve();
return 0;
}