BZOJ 3809 Gty的二逼妹子序列

Description:

有一个长度为n的序列， 有一些询问l r a b，表示区间[l,r]中数权值在[a,b]中的数的种类数。

Solution:

nsqrt(n)logn的很容易想到，但是会超。

考虑莫队时如何快速计算答案？把权值分块，块内统计答案，每次询问只需sqrt(n)。

故总的时间复杂度为nsqrt(n)+msqrt(n)

Code:

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>

using namespace std;

#define REP(i, a, b) for (int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
const int maxn = 1e5+10, maxm = 1e6+10;
int n, m, a[maxn];
int bel[maxn], l[maxn], r[maxn], s_block;
struct Query
{
    int l, r, a, b, id;
    void read(int i) { scanf("%d %d %d %d", &l, &r, &a, &b), id = i; }
    bool operator < (const Query &AI) const { return bel[l] == bel[AI.l] ? r < AI.r : l < AI.l; }
}q[maxm];
int s[1100], c[maxn], ans[maxm];

int calc(int x, int y)
{
    int L = bel[x], R = bel[y], ret = 0;
    if (L == R)
    {
        REP(i, x, y) if (c[i]) ret ++;
    }
    else
    {
        REP(i, L+1, R-1) ret += s[i];
        REP(i, x, r[L]) if (c[i]) ret ++;
        REP(i, l[R], y) if (c[i]) ret ++;
    }
    return ret;
} 

void add(int x) { if (++c[x] == 1) s[bel[x]] ++; }

void del(int x) { if (--c[x] == 0) s[bel[x]] --; }

void solve()
{
    int now_l = 1, now_r = 0;
    REP(i, 1, m)
    {
        for (; now_r < q[i].r; add(a[++now_r]));
        for (; now_l > q[i].l; add(a[--now_l]));
        for (; now_r > q[i].r; del(a[now_r--]));
        for (; now_l < q[i].l; del(a[now_l++]));
        ans[q[i].id] = calc(q[i].a, q[i].b);
    }
}

int main()
{
    scanf("%d %d", &n, &m);
    REP(i, 1, n) scanf("%d", &a[i]);
    int block = int(sqrt(n));
    REP(i, 1, n)
    {
        bel[i] = i/block+1, r[bel[i]] = i;
        if (i == 1 || bel[i] != bel[i-1]) l[bel[i]] = i;
    }
    s_block = n/block+1;
    REP(i, 1, m) q[i].read(i);
    sort(q+1, q+m+1);
    solve();
    REP(i, 1, m) printf("%d
", ans[i]);
    return 0;
}

//if为了省行，少了括号，调了很久