正题

题目链接:https://jzoj.net/senior/#main/show/3859

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题目大意

$n$个台阶，第$i$个高度为$h_i$，把它分为两个集合，使得两个集合中相邻的$h_i$差值和最小。

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解题思路

设$f_i$表示刚好处理到$i$且目前集合开头是$i$的最小差值
定义$$s_i=\sum _{j=1}^i|h_j-h_{j-1}|$$

然后有转移方程fi=min{fj+si−1−sj+∣hi−hj−1∣}(j∈[1..i−1])f_{i}=min\{f_j+s_{i-1}-s_j+|h_i-h_{j-1}|\}(j\in[1..i-1])fi​=min{fj​+si−1​−sj​+∣hi​−hj−1​∣}(j∈[1..i−1])

考虑用数据结构优化上面的$dp$，我们将情况分为：

1. $h_i\ge h_{j-1}:$有转移fi=si−1+hi+min{fj−sj−hj−1}f_i=s_{i-1}+h_i+min\{f_j-s_{j}-h_{j-1}\}fi​=si−1​+hi​+min{fj​−sj​−hj−1​}
2. $h_i\le h_{j-1}:$有转移fi=si−1−hi+min{fj−sj+hj−1}f_i=s_{i-1}-h_i+min\{f_j-s_{j}+h_{j-1}\}fi​=si−1​−hi​+min{fj​−sj​+hj−1​}

用两个树状数组分开维护后面的$min$值即可。

时间复杂度$O(n\log n)$

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define lowbit(x) (x&-x)
using namespace std;
const ll N=5e5+10;
ll n,m,h[N],b[N],s[N],f[N],ans;
struct Tree_Array{ll t[N];void Change(ll x,ll val){while(x<=m){t[x]=min(t[x],val);x+=lowbit(x);}}ll Ask(ll x){ll ans=2147483647;while(x){ans=min(ans,t[x]);x-=lowbit(x);}return ans;}
}Ta,Tp;
int main()
{//freopen("data.txt","r",stdin);memset(Ta.t,0x3f,sizeof(Ta.t));memset(Tp.t,0x3f,sizeof(Tp.t));scanf("%lld",&n);for(ll i=1;i<=n;i++){scanf("%lld",&h[i]);b[i]=h[i];s[i]=s[i-1]+abs(h[i]-h[i-1]);}b[n+1]=0;sort(b+1,b+2+n);m=unique(b+1,b+2+n)-b-1;ans=1e18;ll lp=1,la=m;for(ll i=1;i<=n;i++){ll pre=lower_bound(b+1,b+1+m,h[i])-b;ll aft=m-pre+1;f[i]=min(s[i-1]+h[i]+Tp.Ask(pre),s[i-1]-h[i]+Ta.Ask(aft));if(i==1) f[i]=h[i];Tp.Change(lp,f[i]-s[i]-h[i-1]);Ta.Change(la,f[i]-s[i]+h[i-1]);ans=min(ans,f[i]+s[n]-s[i]);la=aft;lp=pre;}printf("%lld",ans);
}