P2801 分块 + 二分

题意

传送门 P2801 教主的魔法

题解

考虑分块，对于整块的元素，排序后，便可以二分答案；对于左右边界不足一块的元素，朴素修改、统计，并对所在块重新进行排序。设块数为 $t$，由于分块规模与 $N$ 只有指数上的差异，将 $\log (N/t)$ 近似为 $O(\log N)$，那么初始化排序 $O(N\log N)$，单次查询 $O(N/t+t\log N)$，单次修改 $O(N/t+N/t\log N+t)$；当分块大小取 $\sqrt{N}$，满足 $N\approx Q(N/t+t)$。

需要注意的是，排序会修改元素位置，故应保留原始数组；每次排序，将数组复制一份再进行操作。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005, maxt = 1005;
int N, Q, H[maxn], F[maxn];
int tot, id[maxn], L[maxt], R[maxt], add[maxt];

void rst(int i)
{
    
    
    memcpy(F + L[i], H + L[i], sizeof(int) * (R[i] - L[i] + 1));
    sort(F + L[i], F + R[i] + 1);
}

void change(int l, int r, int x)
{
    
    
    int p = id[l], q = id[r];
    if (p == q)
    {
    
    
        for (int i = l; i <= r; ++i)
            H[i] += x;
        rst(p);
    }
    else
    {
    
    
        for (int i = l; i <= R[p]; ++i)
            H[i] += x;
        for (int i = L[q]; i <= r; ++i)
            H[i] += x;
        rst(p), rst(q);
        for (int i = p + 1; i <= q - 1; ++i)
            add[i] += x;
    }
}

int ask(int l, int r, int x)
{
    
    
    int p = id[l], q = id[r], cnt = 0;
    if (p == q)
        for (int i = l; i <= r; ++i)
            cnt += H[i] + add[p] >= x;
    else
    {
    
    
        for (int i = l; i <= R[p]; ++i)
            cnt += H[i] + add[p] >= x;
        for (int i = L[q]; i <= r; ++i)
            cnt += H[i] + add[q] >= x;
        for (int i = p + 1; i <= q - 1; ++i)
            cnt += R[i] - L[i] + 1 - (lower_bound(F + L[i], F + R[i] + 1, x - add[i]) - (F + L[i]));
    }
    return cnt;
}

int main()
{
    
    
    scanf("%d%d", &N, &Q);
    for (int i = 1; i <= N; ++i)
        scanf("%d", H + i);
    memcpy(F + 1, H + 1, sizeof(int) * N);
    int w = sqrt(N);
    for (int i = 1; i <= N; i += w)
    {
    
    
        L[++tot] = i, R[tot] = min(i + w - 1, N);
        sort(F + L[tot], F + R[tot] + 1);
    }
    for (int i = 1; i <= tot; ++i)
        for (int j = L[i]; j <= R[i]; ++j)
            id[j] = i;
    for (int i = 1, l, r, x; i <= Q; ++i)
    {
    
    
        char op;
        scanf(" %c%d%d%d", &op, &l, &r, &x);
        if (op == 'M')
            change(l, r, x);
        else
            printf("%d\n", ask(l, r, x));
    }
}