設f[i][j]為前i個劃成j段的最小代價，枚舉上個劃分點轉移。容易想到這個dp有決策單調性，感性證明一下比較顯然。如果用單調棧維護決策就不太能快速的求出逆序對個數了，改為使用分治，移動端點時樹狀數組維護即可，類似莫隊的每次都在原有基礎上更新。註意更新時先加再減。感覺復雜度非常玄學絲毫不能看出為啥只需要更新nlog次？

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using
 
 namespace std;
#define ll long long
#define N 40010
#define M 11
#define inf 1600000010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    
 
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],f[M][N],tree[N],l,r,cur;
void add(int k,int x){while (k<=n) tree[k]+=x,k+=k&-k;}
int query(int k){int s=0;while (k) s+=tree[k],k-=k&-k;return
 
 s;}
void update(int L,int R)
{
    while (r<R) cur+=r-l+1-query(a[r+1]),add(a[++r],1);
    while (l>L) cur+=query(a[l-1]),add(a[--l],1);
    while (r>R) add(a[r--],-1),cur-=r-l+1-query(a[r+1]);
    while (l<L) add(a[l++],-1),cur-=query(a[l-1]);
}
void solve(int k,int x,int y,int l,int r)
{
    if (l>r) return;
    int mid=l+r>>1,id=min(mid-1,y);f[k][mid]=inf;
    for (int i=min(mid-1,y);i>=x;i--)
    {
        update(i+1,mid);
        if (f[k-1][i]+cur<=f[k][mid]) f[k][mid]=f[k-1][i]+cur,id=i;
    }
    solve(k,x,id,l,mid-1);
    solve(k,id,y,mid+1,r);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5125.in","r",stdin);
    freopen("bzoj5125.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();l=1,r=0;
    for (int i=1;i<=n;i++) a[i]=read();
    f[0][0]=0;for (int i=1;i<=n;i++) f[0][i]=inf;
    for (int j=1;j<=m;j++) solve(j,0,n-1,1,n);
    cout<<f[m][n];
    return 0;
}

BZOJ5125 小Q的書架（決策單調性+動態規劃+分治+樹狀數組）