BZOJ - 3339: Rmq BZOJ - 3585: mex
题解：分块维护权值，用莫队转移。
分块修改操作$O(1)$，查询$O(\sqrt{A_{max}})$。莫队转移$O(m\sqrt n)$。总共是$O(m\sqrt n)$
一份代码解决两道题。额外的经验！
1 #include<cmath> 2 #include<algorithm> 3 #include<cstdio> 4 #include<iostream> 5 using namespace std; 6 inline char nc() { 7 static char b[1<<14],*s=b,*t=b; 8 return s==t&&(t=(s=b)+fread(b,1,1<<14,stdin),s==t)?-1:*s++; 9 } 10 inline void read(int &x) { 11 char b = nc(); x = 0; 12 for (; !isdigit(b); b = nc()); 13 for (; isdigit(b); b = nc()) x = x * 10 + b - '0'; 14 } 15 int n, m, a[200010], ans[200010]; 16 int S, bl[200010], tans, cnt[200010]; 17 struct Q { 18 int l, r, id, t; 19 inline void init(int i) { 20 read(l); read(r); id = i; t = l / S; 21 } 22 inline bool operator<(const Q &q) const { 23 return t < q.t || (t == q.t && r < q.r); 24 } 25 } q[200010]; 26 const int INF = 0x3f3f3f3f; 27 void edt(int x, int k) { 28 if (x > n) return; cnt[x] += k; 29 if (x < tans && !cnt[x]) tans = x; 30 while (cnt[tans]) ++tans; 31 } 32 int main() { 33 read(n); read(m); S = sqrt(n); 34 for (int i = 1; i <= n; ++i) read(a[i]); 35 for (int i = 0; i <= n; ++i) bl[i] = i / S + 1; 36 for (int i = 0; i < m; ++i) q[i].init(i); 37 sort(q, q + m); edt(a[1], 1); 38 for (int l = 1, r = 1, i = 0; i < m; ++i) { 39 while (l < q[i].l) edt(a[l++], -1); 40 while (l > q[i].l) edt(a[--l], 1); 41 while (r < q[i].r) edt(a[++r], 1); 42 while (r > q[i].r) edt(a[r--], -1); 43 ans[q[i].id] = tans; 44 } 45 for (int i = 0; i < m; ++i) printf("%d\n", ans[i]); 46 return 0; 47 }
