题目大意
有\(n\)个数，\(m\)个查询，每次查询一个区间内的逆序对个数。
\(n,m\le 10^5\)
思路
其实是为了锻炼二次离线才做这道题的。
不难想到可以有一个\(\Theta(n\sqrt n\log n)\)的方法，即用莫队，每次用树状数组计算变化的贡献。
然后我们就可以想到二次离线了。我们考虑计算\([l,r]\to [l,r^{'}]\)的贡献，可以发现\([l,r]\to[l,r+1]\)变换的贡献就是\([l,r]\)中比\(a_{r+1}\)大的个数，差分一下，就是\([1,r]\)中比\(a_{r+1}\)大的个数减去\([1,l-1]\)中比\(a_{r+1}\)大的个数。我们发现前面那个东西可以\(\Theta(n\log n)\)预处理出来，后面那个东西可以再次离线下来用值域分块\(\Theta(n\sqrt n)\)求出来。
我们再来考虑\([l,r]\to [l^{'},r]\)的变化。\([l,r]\to [l-1,r]\)的贡献就是\([l,r]\)中比\(a_{l-1}\)小的个数，差分一下，即是\([l,n]\)中比\(a_{l-1}\)小的个数减去\([r+1,n]\)中比\(a_{l-1}\)小的个数。前面那个东西同样可以离线\(\Theta(n\log n)\)，后面那个也可以\(\Theta(n\sqrt n)\)值域分块。
于是，我们的总时间复杂度即为\(\Theta(m\sqrt n+n\sqrt n)\)。
但是这道题非常卡常。
\(\texttt{Code}\)
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define ll long long
#define MAXN 100005
#define MAXM 355
ll ans[MAXN],sum1[MAXN],sum2[MAXN];
int n,m,un,tot,siz,h[MAXN],w[MAXN],bel[MAXN],tmp[MAXN],laz[MAXM],col[MAXM],cor[MAXM],tree[MAXN];
struct Query{
int l,r,id;
bool operator < (const Query &p)const{return (l / siz) != (p.l / siz) ? l < p.l : r < p.r;}
}q[MAXN];
struct node{int p,l,r,id;};
vector <node> vec1[MAXN],vec2[MAXN];
void add1 (int i,int p,int l,int r,int id){vec1[i].push_back (node {p,l,r,id});}
void add2 (int i,int p,int l,int r,int id){vec2[i].push_back (node {p,l,r,id});}
int lowbit (int x){return x & (-x);}
void update (int x,int k){for (;x <= n;x += lowbit (x)) tree[x] += k;}
int query (int x){int sum = 0;for (;x;x -= lowbit (x)) sum += tree[x];return sum;}
void pushup1 (int x){
if (laz[bel[x]]) for (Int i = col[bel[x]];i <= cor[bel[x]];++ i) w[i] += laz[bel[x]];laz[bel[x]] = 0;
for (Int i = col[bel[x]];i <= x;++ i) w[i] ++;
for (Int i = 1;i < bel[x];++ i) laz[i] ++;
}
void pushup2 (int x){
if (laz[bel[x]]) for (Int i = col[bel[x]];i <= cor[bel[x]];++ i) w[i] += laz[bel[x]];laz[bel[x]] = 0;
for (Int i = x;i <= cor[bel[x]];++ i) w[i] ++;
for (Int i = bel[x] + 1;i <= tot;++ i) laz[i] ++;
}
void solve (){//二次离线部分
int cnt = sqrt (un);col[tot = 1] = 1;
if (cnt * cnt < un) ++ cnt;
for (Int i = 1;i <= un;++ i){
bel[i] = tot;
if (i % cnt == 0) cor[tot] = i,col[++ tot] = i + 1;
}
cor[tot] = un;
for (Int i = 1;i <= n;++ i){
for (Int j = 0,len = vec1[i].size();j < len;++ j){
int id = vec1[i][j].id,p = vec1[i][j].p;
for (Int k = vec1[i][j].l;k <= vec1[i][j].r;++ k) ans[id] += 1ll * p * (laz[bel[h[k] + 1]] + w[h[k] + 1]);
}
pushup1 (h[i]);
}
for (Int i = 1;i <= tot;++ i) laz[i] = 0;
for (Int i = 1;i <= un + 1;++ i) w[i] = 0;
for (Int i = n;i >= 1;-- i){
for (Int j = 0,len = vec2[i].size();j < len;++ j){
int id = vec2[i][j].id,p = vec2[i][j].p;
for (Int k = vec2[i][j].l;k <= vec2[i][j].r;++ k) ans[id] += 1ll * p * (laz[bel[h[k] - 1]] + w[h[k] - 1]);
}
pushup2 (h[i]);
}
}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
signed main(){
read (n,m),siz = 317;
for (Int i = 1;i <= n;++ i) read (h[i]),tmp[i] = h[i];
sort (tmp + 1,tmp + n + 1);un = unique (tmp + 1,tmp + n + 1) - tmp - 1;
for (Int i = 1;i <= n;++ i) h[i] = lower_bound (tmp + 1,tmp + un + 1,h[i]) - tmp;
for (Int i = 1;i <= m;++ i) read (q[i].l,q[i].r),q[i].id = i;sort (q + 1,q + m + 1);
for (Int i = 1;i <= n;++ i) sum1[i] = sum1[i - 1] + i - 1 - query (h[i]),update (h[i],1);
for (Int i = 1;i <= n;++ i) tree[i] = 0;
for (Int i = n;i;-- i) sum2[i] = sum2[i + 1] + query (h[i] - 1),update (h[i],1);
int l = 1,r = 0;
for (Int i = 1;i <= m;++ i){
if (r < q[i].r) ans[q[i].id] += sum1[q[i].r] - sum1[r],add1 (l,-1,r + 1,q[i].r,q[i].id),r = q[i].r;
if (r > q[i].r) ans[q[i].id] -= sum1[r] - sum1[q[i].r],add1 (l,1,q[i].r + 1,r,q[i].id),r = q[i].r;
if (l < q[i].l) ans[q[i].id] -= sum2[l] - sum2[q[i].l],add2 (r,1,l,q[i].l - 1,q[i].id),l = q[i].l;
if (l > q[i].l) ans[q[i].id] += sum2[q[i].l] - sum2[l],add2 (r,-1,q[i].l,l - 1,q[i].id),l = q[i].l;
}
solve ();
for (Int i = 1;i <= m;++ i) ans[q[i].id] += ans[q[i - 1].id];
for (Int i = 1;i <= m;++ i) write (ans[i]),putchar ('\n');
return 0;
}