BZOJ - 3339: Rmq BZOJ - 3585: mex

3339: Rmq Problem

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 }

原文地址：https://www.cnblogs.com/p0ny/p/8127849.html