cf1000F. One Occurrence(线段树 set)

题意

题目链接

sol

(真后悔没打这场edu qwq)

首先把询问离线，预处理每个数的\(pre, nxt\)，同时线段树维护\(pre\)(下标是\(pre\)，值是\(i\))，同时维护一下最大值

那么每次在\((1, l - 1)\)内查询最大值，如果最大值\(>= l\)，那么说明合法

但是\(pre\)可能会有相同的情况(0)，直接开个set维护一下

然后用vector对\(nxt\)维护一个类似差分的东西，在\(nxt_i\)的位置删除掉\(i\)的影响

// luogu-judger-enable-o2
/*
*/
#include<bits/stdc++.h> 
#define ll long long
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second 
using namespace std;
const int maxn = 2e6 + 10;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

int n, m, a[maxn], pre[maxn], nxt[maxn], ans[maxn], date[maxn], num = 0;
vector<pair> q[maxn];
void des() {
    for(int i = 1; i <= n; i++) date[i] = a[i];
    sort(date + 1, date + n + 1);
    num = unique(date + 1, date + n + 1) - date - 1;
    for(int i = 1; i <= n; i++) a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date;
}
void get() {
    static int las[maxn];
    for(int i = 1; i <= n; i++) pre[i] = las[a[i]], las[a[i]] = i;
    for(int i = 1; i <= n; i++) las[i] = n + 1;
    for(int i = n; i >= 1; i--) nxt[i] = las[a[i]], las[a[i]] = i;
}
#define getmid ((t[k].l + t[k].r) >> 1)
#define ls k << 1
#define rs k << 1 | 1
struct node {
    int l, r, mx;
}t[maxn];
void update(int k) {
    t[k].mx = max(t[ls].mx, t[rs].mx);
}
void build(int k, int ll, int rr) {
    t[k].l = ll; t[k].r = rr; t[k].mx = 0;
    if(ll == rr) return ;
    int mid = getmid;
    build(ls, ll, mid); build(rs, mid + 1, rr);
}
void modify(int k, int pos, int v) {
    if(t[k].l == t[k].r) {t[k].mx = v; return ;}
    int mid = getmid;
    if(pos <= mid) modify(ls, pos, v);
    if(pos  > mid) modify(rs, pos, v);
    update(k);
}
int query(int k, int ll, int rr) {
    if(ll <= t[k].l && t[k].r <= rr) return t[k].mx;
    int mid = getmid, ans = 0;
    if(ll <= mid) chmax(ans, query(ls, ll, rr));
    if(rr  > mid) chmax(ans, query(rs, ll, rr));
    return ans;
}
#undef ls
#undef rs 
#undef getmid
void solve() {
    set<int> s;
    static vector<int> v[maxn];
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j < v[i].size(); j++) {
            if(!pre[v[i][j]]) 
                s.erase(v[i][j]);
            else modify(1, pre[v[i][j]], 0);
        }
        if(!pre[i]) 
            s.insert(i);
        else modify(1, pre[i], i);
            
        v[nxt[i]].push_back(i);
        for(int j = 0; j < q[i].size(); j++) {
            int t = query(1, 1, q[i][j].fi - 1);
            if(t >= q[i][j].fi) ans[q[i][j].se] = date[a[t]];
            if(!s.empty()) {
                set<int>::iterator it = s.end(); it--;
                if(*it >= q[i][j].fi) ans[q[i][j].se] = date[a[*it]];
            }
            
        }
            
    }
}
signed main() {
    n = read();
    for(int i = 1; i <= n; i++) a[i] = read();
    des();
    get();
    build(1, 1, n + 1);
    m = read();
    for(int i = 1; i <= m; i++) {
        int l = read(), r = read();
        q[r].push_back(mp(l, i));
    }
    solve();
    for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
    return 0;
}
/*
5
1 2 2 1 1
2
1 5
2 3 


10
5 9 6 4 8 7 4 9 7 6
1
4 8
*/