BZOJ 1112: [POI2008]砖块Klo 线段树维护区间中位数

来源：互联网发布：大同市风力发电数据编辑：程序博客网时间：2024/05/01 13:00
时空隧道
翻译一下就是求出任意长度为k的区间的中位数
代码如下：
#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>//by NeighThorn#define inf 100000000000using 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[tr].l=l,tree[tr].r=r,tree[tr].sum=tree[tr].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[tr].l==tree[tr].r){        tree[tr].cnt+=val;tree[tr].sum+=val*pos;        return;    }    int mid=(tree[tr].l+tree[tr].r)>>1;    if(pos<=mid)        change(pos,val,tr<<1);    else        change(pos,val,tr<<1|1);    tree[tr].cnt=tree[tr<<1].cnt+tree[tr<<1|1].cnt;    tree[tr].sum=tree[tr<<1].sum+tree[tr<<1|1].sum;}inline int querynum(int num,int tr){    if(tree[tr].l==tree[tr].r)        return tree[tr].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[tr].l==l&&tree[tr].r==r)        return M(tree[tr].sum,tree[tr].cnt);    int mid=(tree[tr].l+tree[tr].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;}
by >_< NeighThorn
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