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; }转载于:https://www.cnblogs.com/LiM-817/p/10887234.html
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