Luogu P4137 Rmq Problem / mex

题目描述

https://www.luogu.com.cn/problem/P4137

Solution

我们考虑用线段树的经典做法,首先先将操作离线,然后我们枚举左端点,用线段树维护右端点的答案

考虑到左端点右移的时候,会将 $[i,nxt[i]-1]$ 这些位置的数与 $a[i]$ 取最小值,并且最后是单点询问,所以直接那线段树维护即可

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#define maxn 200010
#define pb push_back
#define INF 1000000000
using namespace std;

inline int min(int a, int b, int c) { return min(a, min(b, c)); }

int n, m, a[maxn];

struct node {
int id, p;

node() {}
node(int _id, int _p) { id = _id; p = _p; }
}; vector<node> A[maxn];

bool vis[maxn]; int nxt[maxn], b[maxn];

#define lc i << 1
#define rc i << 1 | 1
struct Seg {
int v, tag;
} T[maxn * 4]; int num[maxn];
void build(int i, int l, int r) {
T[i].tag = INF;
if (l == r) return (void) (T[i].v = num[l]);
int m = l + r >> 1;
build(lc, l, m); build(rc, m + 1, r);
}

void update(int i, int l, int r, int L, int R, int v) {
if (l > R || r < L) return ;
if (L <= l && r <= R) return (void) (T[i].tag = min(T[i].tag, v));
int m = l + r >> 1;
update(lc, l, m, L, R, v); update(rc, m + 1, r, L, R, v);
}

int query(int i, int l, int r, int k, int tag) {
if (l == r) return min(tag, T[i].tag, T[i].v);
int m = l + r >> 1;
if (k <= m) return query(lc, l, m, k, min(T[i].tag, tag));
else return query(rc, m + 1, r, k, min(T[i].tag, tag));
}

int ans[maxn];
int main() {
cin >> n >> m;
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i] = min(n + 1, a[i]), b[a[i]] = n + 1;
for (int i = n; i; --i)
nxt[i] = b[a[i]], b[a[i]] = i;
for (int i = 1; i <= m; ++i) {
int x, y; scanf("%d%d", &x, &y);
A[x].pb(node(i, y));
}
for (int i = 1, now = 0; i <= n; ++i) {
vis[a[i]] = 1;
while (vis[now]) ++now;
num[i] = now;
} build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto u : A[i]) ans[u.id] = query(1, 1, n, u.p, INF);
update(1, 1, n, i, nxt[i] - 1, a[i]);
}
for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
return 0;
}