Bzoj 1112: [poi2008] Brick klo

Description

N-pillar brick, which is expected to have a continuous K-pillar of the same height. you can choose the following two actions: 1. Take a brick from the top of a column and drop it. 2: Pull a brick from the warehouse and put it in another column. the warehouse is infinitely large. now you want to complete the task with a minimum number of actions. input

The first line gives N, K. (1 ≤ k ≤ n ≤ 100000), and the next n rows, each line represents the height of the column brick. 0 ≤ Hi ≤ 1000000

Question:

If [L, R] is determined to be the sameMedianThe minimum number of operations is required.

You can use the Balance Tree for maintenance. You can query the sum of the largest K and smaller K numbers, and the sum of the K and larger K numbers. You can use these to quickly find the answer.

Complexity 'o (nlogn )'. Note ll.

Code:

(Slag written by the Balance Tree .)

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>//by zrt//problem:using namespace std;typedef long long LL;const int inf(0x3f3f3f3f);const double eps(1e-9);LL n,k;LL a[100005];LL num[100005],sum[100005],ls[100005],rs[100005],val[100005],fa[100005],siz[100005];inline void upd(int o){    siz[o]=siz[ls[o]]+siz[rs[o]]+num[o];    sum[o]=num[o]*val[o]+sum[ls[o]]+sum[rs[o]];}inline void zig(int x){    int y=fa[x];    if(rs[x]) ls[y]=rs[x],fa[rs[x]]=y;    else ls[y]=0;    fa[x]=fa[y];    if(fa[y]){        if(ls[fa[y]]==y) ls[fa[y]]=x;else rs[fa[y]]=x;    }    fa[y]=x;rs[x]=y;    upd(y);}inline void zag(int x){    int y=fa[x];    if(ls[x]) rs[y]=ls[x],fa[ls[x]]=y;    else rs[y]=0;    fa[x]=fa[y];    if(fa[y]){        if(ls[fa[y]]==y) ls[fa[y]]=x;else rs[fa[y]]=x;    }    fa[y]=x;ls[x]=y;    upd(y);}int root;inline void splay(int x,int z){    while(fa[x]!=z){        int y=fa[x];        if(fa[y]==z){            if(ls[y]==x) zig(x);            else zag(x);        }else{            if(ls[fa[y]]==y){                if(ls[y]==x){                    zig(y),zig(x);                }else{                    zag(x);zig(x);                }            }else{                if(ls[y]==x){                    zig(x),zag(x);                }else{                    zag(y),zag(x);                }            }        }    }    if(!z) root=x;    upd(x);}int c;inline void insert(LL x){    int o=root;    int f=0;    while(o){        f=o;        if(val[o]==x){            num[o]++;            splay(o,0);            return;        }        if(x<val[o]){            o=ls[o];        }else{            o=rs[o];        }    }    o=++c;    if(f){        if(x<val[f]){            ls[f]=o;        }else{            rs[f]=o;        }    }else root=o;    fa[o]=f;    val[o]=x;    num[o]=1;    splay(o,0);}inline void del(LL x){    int o=root;    while(1){        if(val[o]==x){            num[o]--;            splay(o,0);            return;        }        if(x<val[o]){            o=ls[o];        }else o=rs[o];    }}inline void ask(int k){    int o=root;    while(1){        if(siz[ls[o]]>=k) {            o=ls[o];continue;        }        if(siz[ls[o]]+num[o]>=k) {            splay(o,0);return;        }else{            k-=siz[ls[o]]+num[o];            o=rs[o];        }    }}LL ans;int main(){    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    ans=inf*1LL*1000000;    scanf("%lld%lld",&n,&k);    for(int i=1;i<=n;i++){        scanf("%lld",&a[i]);    }    for(int i=1;i<k;i++){        insert(a[i]);    }    for(int i=k;i<=n;i++){        if(i>k){            del(a[i-k]);        }        insert(a[i]);        ask((k+1)/2);        LL aim=val[root];        ans=min(ans,siz[ls[root]]*aim-sum[ls[root]]+sum[rs[root]]-siz[rs[root]]*aim);    }    printf("%lld\n",ans);    return 0;}

Bzoj 1112: [poi2008] Brick klo