洛谷 5047 Yuno loves sqrt technology II

发现忘了写题解……
来补一份。

二次离线莫队就不讲了。

应用二次离线莫队的时候，会发现我们需要寻找一种数据结构能承受 O(n) 次修改，O(n√n) 次询问。
显然地值域分块，这玩意也可以把修改和询问的复杂度交换，达到一种根号平衡。

就是有点复杂（
听说还有 n1.41 的做法，瑟瑟发抖。

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <algorithm>
#define lowbit(x) ((x) & -(x))
using namespace std;
const int N = 1e5;
const int BLK = 320;
const int CNT = 320;
int n,m,block,a[N + 5],pos[N + 5];
int ind[N + 5],len;
struct s_query
{
    int l,r,id;
    long long ans;
    inline bool operator<(const s_query &o) const
    {
        return pos[l] ^ pos[o.l] ? pos[l] < pos[o.l] : pos[l] & 1 ? r < o.r : r > o.r;
    }
} qry[N + 5];
struct s_contrib
{
    int l,r,id,w;
};
int c[N + 5];
inline void update(int x,int k)
{
    for(;x <= len;x += lowbit(x))
        c[x] += k;
}
inline int query(int x)
{
    int ret = 0;
    for(;x;x -= lowbit(x))
        ret += c[x];
    return ret;
}
int cnt[CNT + 5],pre[N + 5];
inline void add(int x,int k)
{
    for(register int i = pos[x] + 1;i <= pos[n];++i)
        cnt[i] += k;
    for(register int i = x;i <= min(pos[x] * block,n);++i)
        pre[i] += k;
}
inline int ask(int x)
{
    return cnt[pos[x]] + pre[x];
}
vector<s_contrib> vec[2][N + 5];
long long sum[2][N + 5],ans[N + 5];
int main()
{
    scanf("%d%d",&n,&m),block = sqrt(n);
    for(register int i = 1;i <= n;++i)
        scanf("%d",a + i),ind[i] = a[i],pos[i] = (i - 1) / block + 1;
    sort(ind + 1,ind + n + 1),len = unique(ind + 1,ind + n + 1) - ind - 1;
    for(register int i = 1;i <= n;++i)
        a[i] = lower_bound(ind + 1,ind + len + 1,a[i]) - ind;
    for(register int i = 1;i <= n;++i)
        sum[0][i] = i - query(a[i]) - 1,update(a[i],1);
    memset(c,0,sizeof c);
    for(register int i = 1;i <= n;++i)
        sum[1][i] = query(a[i] - 1),update(a[i],1);
    for(register int i = 1;i <= m;++i)
        scanf("%d%d",&qry[i].l,&qry[i].r),qry[i].id = i;
    sort(qry + 1,qry + m + 1);
    for(register int i = 1,l = 1,r = 0;i <= m;++i)
    {
        l < qry[i].l && (vec[1][r].push_back((s_contrib){l,qry[i].l - 1,i,-1}),1);
        for(;l < qry[i].l;qry[i].ans += sum[1][l++]);
        l > qry[i].l && (vec[1][r].push_back((s_contrib){qry[i].l,l - 1,i,1}),1);
        for(;l > qry[i].l;qry[i].ans -= sum[1][--l]);
        r < qry[i].r && (vec[0][l - 1].push_back((s_contrib){r + 1,qry[i].r,i,-1}),1);
        for(;r < qry[i].r;qry[i].ans += sum[0][++r]);
        r > qry[i].r && (vec[0][l - 1].push_back((s_contrib){qry[i].r + 1,r,i,1}),1);
        for(;r > qry[i].r;qry[i].ans -= sum[0][r--]);
    }
    for(register int i = 1;i <= n;++i)
    {
        add(a[i],1);
        for(register vector<s_contrib>::iterator it = vec[0][i].begin();it != vec[0][i].end();++it)
            for(register int j = it->l;j <= it->r;++j)
                qry[it->id].ans += it->w * (i - ask(a[j]));
    }
    memset(cnt,0,sizeof cnt),memset(pre,0,sizeof pre);
    for(register int i = 1;i <= n;++i)
    {
        add(a[i],1);
        for(register vector<s_contrib>::iterator it = vec[1][i].begin();it != vec[1][i].end();++it)
            for(register int j = it->l;j <= it->r;++j)
                qry[it->id].ans += it->w * ask(a[j] - 1);
    }
    for(register int i = 1;i <= m;++i)
        ans[qry[i].id] = (qry[i].ans += qry[i - 1].ans);
    for(register int i = 1;i <= m;++i)
        printf("%lld\n",ans[i]);
}