codeforces 1000F One Occurrence

题意

多次询问lr之间只出现过一次的数是多少。

题解

将查询按照左端点排序，对于所有值维护它在当前位置后面第二次出现是什么时候，那么查询区间最大值即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
//---const int N = 500050;int n, m;
int a[N], ans[N], nx[N], vis[N];
bool in[N];
vector<pii> q[N];struct Seg {
#define ls (rt<<1)
#define rs (ls|1)
static const int N = ::N*4+22;
int ma[N], ind[N];
void upd(int p, int c, int l, int r, int rt) {
if(l == r) {
ma[rt] = c;
ind[rt] = l;
return ;
}
int mid = l+r>>1;
if(p<=mid) upd(p, c, l, mid, ls);
else upd(p, c, mid+1, r, rs);
ma[rt] = max(ma[ls], ma[rs]);
ind[rt] = ma[ls]>ma[rs] ? ind[ls] : ind[rs];
}
void upd(pii &a, pii b) {
if(a < b) a = b;
}
pii qry(int L, int R, int l, int r, int rt) {
if(L<=l && r<=R) return mp(ma[rt], a[ind[rt]]);
int mid = l+r>>1;
pii ans = mp(0, 0);
if(L<=mid) upd(ans, qry(L, R, l, mid, ls));
if(R>=mid+1) upd(ans, qry(L, R, mid+1, r, rs));
return ans;
}
}seg;int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
///read
cin >> n;
rep(i, 1, n+1) cin >> a[i];
cin >> m;
rep(i, 1, m+1) {
int x, y;
cin >> x >> y;
q[x].pb(mp(y, i));
}
///solve
for(int i = n; i; --i) {
nx[i] = vis[a[i]];
if(!nx[i]) nx[i] = n+1;
vis[a[i]] = i;
}
memset(vis, 0, sizeof(vis));
rep(i, 1, n+1) vis[nx[i]] = 1;
rep(i, 1, n+1) if(!vis[i]) seg.upd(i, nx[i], 1, n, 1);
rep(i, 1, n+1) {
for(auto t : q[i]) {
auto c = seg.qry(i, t.fi, 1, n, 1);
ans[t.se] = c.fi>t.fi ? c.se : 0;
}
if(nx[i]<=n) seg.upd(nx[i], nx[nx[i]], 1, n, 1);
}
rep(i, 1, m+1) cout << ans[i] << endl;
return 0;
}