P4137 Rmq Problem / mex (莫队)

题目

p4137 rmq problem / mex

解析

莫队算法维护mex，

• 往里添加数的时候，若添加的数等于\(mex\)，\(mex\)就不能等于这个值了，就从这个数开始枚举找\(mex\)；若不等于\(mex\)，没有影响.

• 取出数的时候，如果这个数出现的次数变为了\(0\)，\(mex\)就和这个数取一个\(min\)

  代码

#include <bits/stdc++.h>
using namespace std;
const int n = 1e6 + 10;
int n, m, mex;
int a[n], cnt[n], ans[n];
class node {
    public :
        int l, r, id, bl;
        bool operator < (const node &oth) const {
            return this->bl == oth.bl ? this->r < oth.r : this->l < oth.l;
        }
} e[n];

template<class t>inline void read(t &x) {
    x = 0; int f = 0; char ch = getchar();
    while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    x = f ? -x : x; 
    return ;
} 

inline void add(int x) {
    cnt[a[x]]++;
    int i = mex;
    if (a[x] == mex) while (cnt[i]) i++;
    mex = i;
}

inline void del(int x) {
    cnt[a[x]]--;
    if (cnt[a[x]] == 0) mex = min(mex, a[x]);
}

int main() {
    read(n), read(m);
    int k = sqrt(n);
    for (int i = 1; i <= n; ++i) read(a[i]);
    for (int i = 1, x, y; i <= m; ++i) {
        read(x), read(y);
        e[i] = (node) {x, y, i, x / k + 1};
    }
    sort(e + 1, e + 1 + m);
    int l = 1, r = 0;
    for (int i = 1; i <= m; ++i) {
        int ll = e[i].l, rr = e[i].r;
        while (l < ll) del(l++);
        while (l > ll) add(--l);
        while (r < rr) add(++r);
        while (r > rr) del(r--);
        ans[e[i].id] = mex;
    }
    for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
    return 0;
}