[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 <<"
";
			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
", ans[i]);
	return 0;
}