luogu P5047 [Ynoi2019 模拟赛] Yuno loves sqrt technology II

https://www.luogu.com.cn/problem/P5047

静态区间逆序对，还卡空间，一看就是根号算法
$O(n\sqrt{n}logn$用莫队+树状数组的做法可以轻松得到，但是带个log貌似过不了

考虑二次离线莫队

考虑每次移动的增量

• $l$向左移动:（向右就是符号取反罢了）
  $\sum (r,a_l-1)-(l,a_l-1)$
• $r$向右移动
  $\sum (r-l)-((r-1,a_r)-(l-1,a_r))$
  把有关$a$的分成四部分考虑
  $$A:(i,a_i-1),B:(i,a_i-1),C:(x,a_i-1),D:(x,a_i)$$
  然后把上式子对应的部分替代
  $A,B$是可以预处理的
  把莫队的$C,D$再离线下来挂到对应位置
  可以发现一定是一段区间，可以只用记录两个端点
  然后再用值域分块来做
  具体看代码吧
  code:

#include<bits/stdc++.h>
#define N 100050
#define ll long long
#define lowbit(x) (x & -x)
using namespace std;
int t[N];
void update(int x, int y) {
    for(; x < N; x += lowbit(x)) t[x] += y;
}
int query(int x) {
    int ret = 0;
    for(; x; x -= lowbit(x)) ret += t[x];
    return ret;
}

int bel[N], n, m, blo;
ll sa[N], sb[N], ans[N];

struct Q {
    int l, r, id;
} q[N];

int cmp(Q x, Q y) {
    return bel[x.l] == bel[y.l]? x.r < y.r : x.l < y.l;
}
struct A {
    int l, r, py, o, id;
};
vector<A> qu[N];
void modui() {
    sort(q + 1, q + 1 + m, cmp);
    int l = 1, r = 0;
    for(int i = 1; i <= m; i ++) {
        if(q[i].l < l) {
            ans[q[i].id] -= sa[l - 1] - sa[q[i].l - 1];
            qu[r].push_back((A){q[i].l, l - 1, - 1, 1, q[i].id});//把C，D离线下来
            l = q[i].l;
        }
        if(r < q[i].r) {
            ans[q[i].id] += 1ll * (r + 1 + q[i].r - 2 * l) * (q[i].r - r) / 2 - (sb[q[i].r] - sb[r]);
       //     printf("** %lld\n",  1ll * (r + 1 + q[i].r - 2 * l) * (q[i].r - r) / 2);
            qu[l - 1].push_back((A){r + 1, q[i].r, 0, 1, q[i].id});
            r = q[i].r;
        }
        if(q[i].l > l) {
            ans[q[i].id] += sa[q[i].l - 1] - sa[l - 1];
            qu[r].push_back((A){l, q[i].l - 1, - 1, - 1, q[i].id});
            l = q[i].l;
        }
        if(r > q[i].r) {
            ans[q[i].id] -= 1ll * (r + 1 + q[i].r - 2 * l) * (r - q[i].r) / 2 - (sb[r] - sb[q[i].r]);
            qu[l - 1].push_back((A){q[i].r + 1, r, 0, - 1, q[i].id});
            r = q[i].r;
        }
      //  printf("  %d %d\n", l, r);
    }
}
struct FK {
    int a[N], tg[N];
    void add(int x) {
        int id = bel[x];
        for(int i = id + 1; i <= bel[n]; i ++) tg[i] ++;
        for(int i = x; i <= id * blo; i ++) a[i] ++;
    }
    int query(int x) {
        return a[x] + tg[bel[x]];
    }
} S;
int a[N], b[N];
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + 1 + n);
    for(int i = 1; i <= n; i ++) a[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
    
    blo = 2 * m / (int)sqrt(n) + 1;
    for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / blo + 1;
    
    for(int i = 1; i <= n; i ++) {//预处理A，B
        sb[i] = sb[i - 1] + query(a[i]);
        update(a[i], 1);
        sa[i] = sa[i - 1] + query(a[i] - 1);
    }
  //  for(int i = 1; i <= n; i ++) printf("%lld ", sb[i]); printf("\n");
    for(int i = 1; i <= m; i ++) scanf("%d%d", &q[i].l, &q[i].r), q[i].id = i;
    modui();
    
    for(int i = 1; i <= n; i ++) {
        S.add(a[i]);
        for(int j = 0; j < qu[i].size(); j ++) {
            A x = qu[i][j];
          //  printf("%d %d %d %d\n", x.id, x.l, x.r, x.py);
            for(int k = x.l; k <= x.r; k ++) {
                ans[x.id] += x.o * S.query(a[k] + x.py);//计算CD
            }
        }
    }
    for(int i = 1; i <= m; i ++) ans[q[i].id] += ans[q[i - 1].id];//因为我们之前计算的是增量，所以这里要做一个前缀和
    for(int i = 1; i <= m; i ++) printf("%lld\n", ans[i]);
    return 0;
}
/*
 5 5
 1 4 2 3 2
 2 4
 3 4
 4 4
 1 4
 1 5
 */