BZOJ 1112: [POI2008]砖块Klo Splay + 性质分析

Code:

#include<bits/stdc++.h>
using namespace std;
#define setIO(s) freopen(s".in","r",stdin)
#define maxn 200000
#define inf 10000000000000
#define get(x) (ch[f[x]][1]==x)
#define ll long long
int root,cc,kk;
int h[maxn],ch[maxn][2],f[maxn],siz[maxn],val[maxn];
long long sumv[maxn];
long long ans=inf;
void pushup(int x)
{
    sumv[x]=sumv[ch[x][0]]+sumv[ch[x][1]]+h[x];
    siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void rotate(int x)
{
    int old=f[x],fold=f[old],which=get(x);
    ch[old][which]=ch[x][which^1],f[ch[old][which]]=old;
    ch[x][which^1]=old,f[old]=x,f[x]=fold;
    if(fold) ch[fold][ch[fold][1]==old]=x;
    pushup(x),pushup(old);
}
void splay(int x,int &tar)
{
    int u=f[tar];
    for(int fa;(fa=f[x])!=u;rotate(x))
        if(f[fa]!=u)
            rotate(get(fa)==get(x)?fa:x);
    tar=x;
}
void insert(int &x,int key,int ff)
{
    if(!x) x=key,f[x]=ff;
    else insert(ch[x][h[key]>h[x]],key,x);
    pushup(x);
}
int query(int kth)
{
    int x=root;
    while(1)
    {
        if(siz[ch[x][0]]+1==kth) return x;
        if(siz[ch[x][0]]>=kth) x=ch[x][0];
        else kth-=(siz[ch[x][0]]+1),x=ch[x][1];
    }
}
void del(int x)
{
    if(!ch[x][0]) root=ch[x][1], f[root]=ch[x][1]=0;
    else if(!ch[x][1]) root=ch[x][0], f[root]=ch[x][0]=0;
    else
    {
        int l=ch[x][0];
        while(ch[l][1]) l=ch[l][1];
        splay(l, ch[x][0]);
        ch[l][1]=ch[x][1], f[ch[x][1]]=l, f[l]=ch[x][0]=ch[x][1]=0, pushup(l);
        root=l;
    }
}
void solve(int k,int L)
{
    int mid=(k%2==0)?k/2:(k/2)+1,x=query(mid);
    splay(x,root);
    int l=ch[root][0],r=ch[root][1];
    ll re=0;
    re+=(h[x]*siz[l]-sumv[l]);
    re+=(sumv[r]-h[x]*siz[r]);
    if(re<ans) ans=re, cc=L, kk=h[x];
}
int main()
{
    // setIO("input");
    int n,k,i,j;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;++i) scanf("%d",&h[i]);
    for(i=1;i<=k;++i) { insert(root,i,0); if(i%6==0) splay(i,root); }
    solve(k,1);
    for(i=k+1;i<=n;++i)
    {
        j=i-k+1;
        splay(j-1, root), del(j-1), insert(root,i,0), splay(i,root),  solve(k,j);
    }
    printf("%lld\n",ans);
    return 0;
}