BZOJ5125 小Q的书架（决策单调性+动态规划+分治+树状数组）

　　设f[i][j]为前i个划成j段的最小代价，枚举上个划分点转移。容易想到这个dp有决策单调性，感性证明一下比较显然。如果用单调栈维护决策就不太能快速的求出逆序对个数了，改为使用分治，移动端点时树状数组维护即可，类似莫队的每次都在原有基础上更新。注意更新时先加再减。感觉复杂度非常玄学丝毫不能看出为啥只需要更新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<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),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=;while (k) s+=tree[k],k-=k&-k;return s;}
void update(int L,int R)
{
    while (r<R) cur+=r-l+-query(a[r+]),add(a[++r],);
    while (l>L) cur+=query(a[l-]),add(a[--l],);
    while (r>R) add(a[r--],-),cur-=r-l+-query(a[r+]);
    while (l<L) add(a[l++],-),cur-=query(a[l-]);
}
void solve(int k,int x,int y,int l,int r)
{
    if (l>r) return;
    int mid=l+r>>,id=min(mid-,y);f[k][mid]=inf;
    for (int i=min(mid-,y);i>=x;i--)
    {
        update(i+,mid);
        if (f[k-][i]+cur<=f[k][mid]) f[k][mid]=f[k-][i]+cur,id=i;
    }
    solve(k,x,id,l,mid-);
    solve(k,id,y,mid+,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=,r=;
    for (int i=;i<=n;i++) a[i]=read();
    f[][]=;for (int i=;i<=n;i++) f[][i]=inf;
    for (int j=;j<=m;j++) solve(j,,n-,,n);
    cout<<f[m][n];
    return ;
}