POJ2104 区间第k小

题意就是区间第k大……

题解：

前段时间用主席树搞掉了……

如今看到划分树，是在想来写一遍，结果18号对着学长的代码调了一上午连样例都没过，好桑心……

今天在做NOI2010超级钢琴，忽然发现用划分树很直观，果断决定再战划分树

对着网上的c++代码抄了一遍，A了，可是这编程复杂度有点高，忽然又看见盾哥的代码

很简短，和我原先的代码差不多，他怎么能A了呢……（后来发现区间第k小，我却写的第k大……sb啊）

考虑到盾哥的程序用到了离散化，因此没有考虑存在两个数相等的情况，这样就使代码减少很多

我报着试一试的心态把我认为肯定对的代码随机生成了几组大数据和盾哥的程序对拍，然后就对上了

好的，以后就用盾哥的程序

代码：

c++翻译来的：

var  s,t:array[0..20,0..200000] of longint;
     a,rk:array[0..200000] of longint;
     i,n,m,x,y,k,j:longint;
 procedure sort(l,r:longint);
 var i,j,m,temp:longint;
 begin
 i:=l;j:=r;m:=a[(i+j)>>1];
 repeat
  while a[i]<m do inc(i);
  while a[j]>m do dec(j);
  if i<=j then
   begin
   temp:=a[i];a[i]:=a[j];a[j]:=temp;
   inc(i);dec(j);
   end;
 until i>j;
 if i<r then sort(i,r);
 if j>l then sort(l,j);
 end;
  procedure init;
   begin
     readln(n,m);
     for i:=1 to n do read(a[i]);t[0]:=a;
     sort(1,n);
   end;
 procedure build(h,l,r:longint);
  var  mid,i,lp,rp,lm:longint;
  begin
  mid:=(l+r)>>1;lm:=mid-l+1;lp:=l;rp:=mid+1;
  for i:=l to mid do
   if a[i]<a[mid] then dec(lm);
  for i:=l to r do
   begin
   if i=l then s[h,i]:=0 else s[h,i]:=s[h,i-1];
   if t[h,i]=a[mid] then
    begin
    if lm<>0 then
     begin
     dec(lm);inc(s[h,i]);t[h+1,lp]:=t[h,i];inc(lp);
     end
    else begin t[h+1,rp]:=t[h,i];inc(rp);end;
    end
   else
    if t[h,i]<a[mid] then begin inc(s[h,i]);t[h+1,lp]:=t[h,i];inc(lp);end
   else begin t[h+1,rp]:=t[h,i];inc(rp);end;
   end;
  if l=r then exit;
  build(h+1,l,mid);
  build(h+1,mid+1,r);
  end;
 function find(h,l,r,x,y,k:longint):longint;
  var mid,ll,rr:longint;
  begin
  if l=r then exit(t[h,l]);
  mid:=(l+r)>>1;
  if l=x then ll:=0 else ll:=s[h,x-1];rr:=s[h,y]-ll;
  if rr>=k then exit(find(h+1,l,mid,l+ll,l+ll+rr-1,k))
  else exit(find(h+1,mid+1,r,mid+1+x-l-ll,mid+1+y-l-ll-rr,k-rr));
  end;
 procedure main;
  begin
   build(0,1,n);
   for i:=1 to m do
    begin
     readln(x,y,k);
     writeln(find(0,1,n,x,y,k));
    end;
  end;
begin
  init;
  main;
end.       

盾哥的：

program poj2104; const maxn=200000;
var
  a,b:array[0..maxn] of longint;
  s,d:array[1..20,0..maxn] of longint;
  n,m,i,j,x,y,z,ls,rs,mid,p:longint;

procedure sort(l,r:longint); var i,j,x,y:longint;
begin
  i:=l;j:=r;x:=a[(i+j)>> 1];
  repeat
    while b[a[i]]<b[x] do inc(i);
    while b[a[j]]>b[x] do dec(j);
    if i<=j then begin
      y:=a[i];a[i]:=a[j];a[j]:=y;
      inc(i);dec(j);
    end;
  until i>j;
  if l<j then sort(l,j);
  if i<r then sort(i,r);
end;

procedure buildtree(h,l,r:longint);var mid:longint;
begin
  if l=r then exit;
  mid:=(l+r+1)>> 1;p:=0;
  for i:=l to r do
    if d[h,i]<mid then begin
      d[h+1,l+p]:=d[h,i];
      inc(p);s[h,i]:=p;
    end else begin
      d[h+1,mid+i-l-p]:=d[h,i];
      s[h,i]:=p;
    end;
  buildtree(h+1,l,mid-1);
  buildtree(h+1,mid,r);
end;

function find(h,ll,rr,l,r,k:longint):longint;
begin
  if l=r then exit(d[h,r]);
  if l=ll then ls:=0 else ls:=s[h,l-1];
  rs:=s[h,r];mid:=(ll+rr+1)>> 1;
  if rs-ls<k
    then find:=find(h+1,mid,rr,l-ll-ls+mid,r-ll-rs+mid,k-rs+ls)
    else find:=find(h+1,ll,mid-1,ls+ll,rs+ll-1,k);
end;

begin
  readln(n,m);
  for i:=1 to n do begin read(b[i]);a[i]:=i; end;
  sort(1,n);
  for i:=1 to n do d[1,a[i]]:=i;
  buildtree(1,1,n);
  for i:=1 to m do begin
    readln(x,y,z);
    writeln(b[a[find(1,1,n,x,y,z)]]);
  end;
end.   

我最先抄的：

const maxn=100000+10;
type arrtype=array[0..maxn] of longint;
var s,z:array[0..20,0..maxn] of longint;
    a,b,rk:arrtype;
    i,j,n,m,x,y,k:longint;
procedure sort(var a:arrtype;l,r:longint);
 var i,j,m,temp:longint;
 begin
 i:=l;j:=r;m:=a[(i+j)>>1];
 repeat
  while a[i]<m do inc(i);
  while a[j]>m do dec(j);
  if i<=j then
   begin
   temp:=a[i];a[i]:=a[j];a[j]:=temp;
   temp:=rk[i];rk[i]:=rk[j];rk[j]:=temp;
   inc(i);dec(j);
   end;
 until i>j;
 if i<r then sort(a,i,r);
 if j>l then sort(a,l,j);
 end;
procedure init;
 begin
 readln(n,m);
 for i:=1 to n do read(a[i]);readln;b:=a;
 for i:=1 to n do rk[i]:=i;
 sort(a,1,n);
 end;
procedure build(h,l,r:longint);
 var i,p,mid:longint;
 begin
 mid:=(l+r)>>1;p:=0;
 for i:=l to r do
  if z[h,i]<=mid then
   begin
    z[h+1,l+p]:=z[h,i];
    inc(p);
    s[h,i]:=p;
   end
  else
   begin
    z[h+1,mid+1+i-p-l]:=z[h,i];
    s[h,i]:=p;
   end;
 if l=r then exit;
 build(h+1,l,mid);
 build(h+1,mid+1,r);
 end;
function find(h,l,r,x,y,k:longint):longint;
 var ll,rr,mid:longint;
 begin
 if l=r then exit(z[h,l]);
 mid:=(l+r)>>1;
 if l=x then ll:=0 else ll:=s[h,x-1];rr:=s[h,y];
 if rr-ll>=k then exit(find(h+1,l,mid,l+ll,l+rr-1,k))
 else exit(find(h+1,mid+1,r,mid+1+x-l-ll,mid+1+y-l-rr,k-(rr-ll)));
 end;
procedure main;
 begin
 for i:=1 to n do z[1,rk[i]]:=i;
 build(1,1,n);
 for i:=1 to m do
  begin
  readln(x,y,k);
  writeln(b[rk[find(1,1,n,x,y,k)]]);
  end;
 end;
begin
 init;
 main;
end.        

盾哥的代码风格确实比我好很多，值得学习

UPD:

c++

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#define inf 1000000000 
#define maxn 100000+100
#define maxm 100000
using namespace std;
int n,m,a[maxn],sa[maxn],t[20][maxn],s[20][maxn];
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void build(int h,int l,int r)
{
    int mid=(l+r)>>1,num=0;
    for(int i=l;i<=r;i++)
    if (t[h][i]<=mid) 
    {
        t[h+1][l+num]=t[h][i];
        s[h][i]=++num;
    }
    else
    {
        t[h+1][mid+1+i-l-num]=t[h][i];
        s[h][i]=num;
    }
    if(l==r)return;
    build(h+1,l,mid);
    build(h+1,mid+1,r);
}
int find(int h,int l,int r,int x,int y,int k)
{
    if(l==r)return t[h][l];
    int mid=(l+r)>>1,ll,rr;
    ll=(l==x)?0:s[h][x-1];rr=s[h][y];
    if(rr-ll>=k) return find(h+1,l,mid,l+ll,l+rr-1,k);
    else return find(h+1,mid+1,r,mid+1+x-l-ll,mid+1+y-l-rr,k-(rr-ll));
}
bool cmp(int x,int y)
{
    return a[x]<a[y];
} 
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=read();m=read();
    for(int i=1;i<=n;i++){a[i]=read();sa[i]=i;}
    sort(sa+1,sa+n+1,cmp);
    for(int i=1;i<=n;i++)t[0][sa[i]]=i;
    build(0,1,n);
    for(int i=1;i<=m;i++)
    {
        int x,y,k;
        x=read();y=read();k=read();
        printf("%d\n",a[sa[find(0,1,n,x,y,k)]]);
    }
    return 0;
}