BZOJ 1112: [POI2008]砖块Klo 线段树维护区间中位数

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
#define inf 100000000000
using namespace std;
const int maxn=100000+5,maxh=1000000+5;
int n,k,h[maxn],m,midnum;
long long ans;
struct Tree{
int l,r,cnt;
long long sum;
}tree[maxh*4];
struct M{
int cnt;
long long sum;
M(long long x=0,int y=0){
sum=x,cnt=y;
}
}aql,anqila;
inline int read(void){
char ch=getchar();int x=0;
while(!(ch>='0'&&ch<='9'))
ch=getchar();
while(ch>='0'&&ch<='9')
x=x*10+ch-'0',ch=getchar();
return x;
}
inline void build(int l,int r,int tr){
tree.l=l,tree.r=r,tree.sum=tree.cnt=0;
if(l==r)
return;
int mid=(l+r)>>1;
build(l,mid,tr<<1);build(mid+1,r,tr<<1|1);
}
inline void change(int pos,int val,int tr){
if(tree.l==tree.r){
tree.cnt+=val;tree.sum+=val*pos;
return;
}
int mid=(tree.l+tree.r)>>1;
if(pos<=mid)
change(pos,val,tr<<1);
else
change(pos,val,tr<<1|1);
tree.cnt=tree[tr<<1].cnt+tree[tr<<1|1].cnt;
tree.sum=tree[tr<<1].sum+tree[tr<<1|1].sum;
}
inline int querynum(int num,int tr){
if(tree.l==tree.r)
return tree.l;
if(tree[tr<<1].cnt<num)
return querynum(num-tree[tr<<1].cnt,tr<<1|1);
else
return querynum(num,tr<<1);
}
inline M querysum(int l,int r,int tr){
if(l<-1)
return M(0,0);
if(l>r)
return M(0,0);
if(tree.l==l&&tree.r==r)
return M(tree.sum,tree.cnt);
int mid=(tree.l+tree.r)>>1;
if(r<=mid)
return querysum(l,r,tr<<1);
else if(l>mid)
return querysum(l,r,tr<<1|1);
else{
M tmp=querysum(l,mid,tr<<1),temp=querysum(mid+1,r,tr<<1|1);
tmp.cnt+=temp.cnt,tmp.sum+=temp.sum;
return tmp;
}
}
signed main(void){
n=read();k=read();m=0;
for(int i=1;i<=n;i++)
h[i]=read(),m=max(m,h[i]);
if(k==1){
puts("0");return 0;
}
build(0,m,1);midnum=(1+k)>>1;
for(int i=1;i<=k;i++)
change(h[i],1,1);
int lala=querynum(midnum,1);aql=querysum(0,lala-1,1),anqila=querysum(lala+1,m,1);ans=aql.cnt*lala-aql.sum+anqila.sum-lala*anqila.cnt;
//  cout<<"lala"<<lala<<" "<<ans<<endl;
for(int i=k+1;i<=n;i++)
change(h[i-k],-1,1),change(h[i],1,1),lala=querynum(midnum,1),/*cout<<"lala"<<lala<<" ",*/aql=querysum(0,lala-1,1),anqila=querysum(lala+1,m,1)/*,cout<<aql.cnt<<" "<<aql.sum<<" "<<anqila.cnt<<" "<<anqila.sum<<endl*/,ans=min(ans,aql.cnt*lala-aql.sum+anqila.sum-lala*anqila.cnt)/*,cout<<ans<<endl*/;
printf("%lld\n",ans);
return 0;
}