[CF1000F]One Occurrence 题解

一句话题意：给定一个数列，每次询问一段区间内有没有只出现一次的数，如果有随便输出一个，否则输出0.

维护last[i]，就是前一个a[i]的位置. 如果是第一个出现，last[i] = 0.
然后对于每一个数，类似于HH的项链一题的做法，将i位置改成last[i], last[i]位置改成inf。这样区间查询的最小值只要<l就有。
搞一颗单点修改区间查询的线段树就可以了。

#include <bits/stdc++.h>
#define ls(p) (p<<1)
#define rs(p) (p<<1|1)
#define fi first
#define se second
using namespace std;
const int N = 500000 + 5;
const int inf = 0x3f3f3f3f;
int n, m;
vector <pair<int , int> > q[N];
int pre[N], tmp[N], a[N];
int mn[N<<2], mnp[N<<2], ans[N<<2];
void Pushup(int p) {
    if(mn[ls(p)] < mn[rs(p)]) {mn[p] = mn[ls(p)]; mnp[p] = mnp[ls(p)];}
    else {mn[p] = mn[rs(p)]; mnp[p] = mnp[rs(p)];}
}
void Build(int p, int l, int r) {
    if(l == r) {
        mn[p] = inf;
        mnp[p] = a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    Build(ls(p), l, mid);
    Build(rs(p), mid + 1, r);
    Pushup(p);
}
void Update(int p, int x, int y, int l, int r) {
    if(l == r) {
        mn[p] = y;
        return;
    }
    if(x < l || x > r) return ;
    int mid = (l + r) >> 1;
    if(x <= mid) Update(ls(p), x, y, l, mid);
    else Update(rs(p), x, y, mid + 1, r);
    Pushup(p);
}

pair<int, int> Query(int p, int ql, int qr, int l, int r) {
    if(ql <= l && r <= qr) return {mn[p], mnp[p]};
    int mid = (l + r) >> 1; pair <int, int> ret = {inf, 0};
    if(ql <= mid) ret = min(ret, Query(ls(p), ql, qr, l, mid));
    if(qr > mid) ret = min(ret, Query(rs(p), ql, qr, mid + 1, r));
    return ret;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), pre[i] = tmp[a[i]], tmp[a[i]] = i;
    //for(int i = 1; i <= n; i++) cout << pre[i] << " ";
    //cout <<endl;
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        int l, r; scanf("%d%d", &l, &r);
        q[r].push_back({l, i});
    }

    Build(1, 1, n);
    for(int i = 1; i <= n; i++) {
        Update(1, i, pre[i], 1, n);
        if(pre[i]) Update(1, pre[i], inf, 1, n);
        for(int j = 0; j < q[i].size(); j++) {
            int l = q[i][j].fi, id = q[i][j].se;
        //cout << "searching query " << l << " " << i <<"\n";
            pair <int, int> node = Query(1, l, i, 1, n);
            if(node.fi < l) {
                ans[id] = node.se;
            }
            else ans[id] = 0;
        }
    }
    for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
    return 0;
}