PKU2823 sliding window

       这是一道经典的数据结构题，可以用来为多种数据结构入门或者热身。它涉及到的数据结构有单调队列、堆结构、线段树。后两项应该是大家比较熟悉的，所以今天做讲解的是单调队列。

       我们以最小值的情况为例。对于两个元素a[i]和a[j]，我们假设i<j。则在本题中一定有a[i]在a[j]之前被删除。如果又有a[j]<a[i]那么a[i]便一定不是最优的，所以可以提前删除a[i]这个无用元素。

       根据均摊思想，单调队列的时间复杂度为O(n)，应该是最优的。

       具体比对请见代码！

 

【单调队列】

 

program sliding_window;var nummax,nummin,quemax,quemin,ansmax,ansmin:array[-5..1000005] of longint; n,m,headmax,tailmax,headmin,tailmin,sum:longint;procedure init;begin readln(n,m);end;procedure prework;begin headmax:=1; headmin:=1; tailmax:=0; tailmin:=0; sum:=0;end;procedure makeque(i:longint);var w:longint;begin read(w); while (headmax<=tailmax) and (quemax[tailmax]<=w) do dec(tailmax); while (headmin<=tailmin) and (quemin[tailmin]>=w) do dec(tailmin); inc(tailmax); quemax[tailmax]:=w; nummax[tailmax]:=i; inc(tailmin); quemin[tailmin]:=w; nummin[tailmin]:=i;end;procedure main;var i:longint;begin for i:=1 to m-1 do makeque(i); for i:=m to n do begin makeque(i); inc(sum); ansmax[sum]:=quemax[headmax]; ansmin[sum]:=quemin[headmin]; while (headmax<=tailmax) and (nummax[headmax]<=sum) do inc(headmax); while (headmin<=tailmin) and (nummin[headmin]<=sum) do inc(headmin); end;end;procedure outit;var i:longint;begin for i:=1 to sum-1 do write(ansmin[i],' '); writeln(ansmin[sum]); for i:=1 to sum-1 do write(ansmax[i],' '); writeln(ansmax[sum]);end;begin assign(input,'window.in'); reset(input); assign(output,'window.out'); rewrite(output); init; prework; main; outit; close(input); close(output);end.

 

【堆结构】

 

 

program sliding_window;type arr=array[0..1000000] of longint;var data,ansmin,ansmax,heapmin,heapmax:arr; n,m,summin,summax,tot:longint;procedure init;var i:longint;begin readln(n,m); for i:=1 to n do read(data[i]);end;procedure prework;begin tot:=0; summin:=0; summax:=0;end;procedure insertmin(x:longint;var n:longint);var i,j:longint;begin inc(n); i:=n; while i<>1 do begin j:=i shr 1; if data[heapmin[j]]>data[x] then begin heapmin[i]:=heapmin[j]; i:=j; end else break; end; heapmin[i]:=x;end;procedure insertmax(x:longint;var n:longint);var i,j:longint;begin inc(n); i:=n; while i<>1 do begin j:=i shr 1; if data[heapmax[j]]<data[x] then begin heapmax[i]:=heapmax[j]; i:=j; end else break; end; heapmax[i]:=x;end;procedure deletemin(i:longint;var n:longint);var j,x:longint;begin x:=heapmin[n]; dec(n); while i shl 1<=n do begin j:=i shl 1; if (j+1<=n) and (data[heapmin[j+1]]<data[heapmin[j]]) then inc(j); if data[x]>data[heapmin[j]] then begin heapmin[i]:=heapmin[j]; i:=j; end else break; end; heapmin[i]:=x;end;procedure deletemax(i:longint;var n:longint);var j,x:longint;begin x:=heapmax[n]; dec(n); while i shl 1<=n do begin j:=i shl 1; if (j+1<=n) and (data[heapmax[j+1]]>data[heapmax[j]]) then inc(j); if data[x]<data[heapmax[j]] then begin heapmax[i]:=heapmax[j]; i:=j; end else break; end; heapmax[i]:=x;end;procedure main;var i:longint;begin for i:=1 to m-1 do begin insertmin(i,summin); insertmax(i,summax); end; for i:=m to n do begin insertmin(i,summin); insertmax(i,summax); inc(tot); while (summin>=1) and (heapmin[1]<tot) do deletemin(1,summin); while (summax>=1) and (heapmax[1]<tot) do deletemax(1,summax); ansmin[tot]:=data[heapmin[1]]; ansmax[tot]:=data[heapmax[1]]; end;end;procedure outit;var i:longint;begin for i:=1 to tot-1 do write(ansmin[i],' '); writeln(ansmin[tot]); for i:=1 to tot-1 do write(ansmax[i],' '); writeln(ansmax[tot]);end;begin assign(input,'window.in'); reset(input); assign(output,'window.out'); rewrite(output); init; prework; main; outit; close(input); close(output);end.

 

【线段树】

 

 

 program sliding_window;type rec=record left,right,min,max:longint; end;var tree:array[0..4000000] of rec; ansmin,ansmax,data:array[0..1000000] of longint; n,m:longint;procedure init;var i:longint;begin readln(n,m); for i:=1 to n do read(data[i]);end;function getmin(a,b:longint):longint;begin if a<b then exit(a); exit(b);end;function getmax(a,b:longint):longint;begin if a>b then exit(a); exit(b);end;procedure buildtree(i,l,r:longint);var m,a,b:longint;begin with tree[i] do begin left:=l; right:=r; if l=r then begin min:=data[l]; max:=data[l]; exit; end; m:=(left+right) shr 1; a:=i shl 1; b:=i shl 1+1; buildtree(a,l,m); buildtree(b,m+1,r); min:=getmin(tree[a].min,tree[b].min); max:=getmax(tree[a].max,tree[b].max); end;end;function askmin(i,l,r:longint):longint;var m,a,b:longint;begin with tree[i] do begin if (l=left) and (r=right) then exit(min); m:=(left+right) shr 1; a:=i shl 1; b:=i shl 1 + 1; if r<=m then exit(askmin(a,l,r)) else if l>=m+1 then exit(askmin(b,l,r)) else exit(getmin(askmin(a,l,m),askmin(b,m+1,r))); end;end;function askmax(i,l,r:longint):longint;var m,a,b:longint;begin with tree[i] do begin if (l=left) and (r=right) then exit(max); m:=(left+right) shr 1; a:=i shl 1; b:=i shl 1 + 1; if r<=m then exit(askmax(a,l,r)) else if l>=m+1 then exit(askmax(b,l,r)) else exit(getmax(askmax(a,l,m),askmax(b,m+1,r))); end;end;procedure main;var l,r,i:longint;begin for i:=1 to n-m+1 do begin l:=i; r:=i+m-1; ansmin[i]:=askmin(1,l,r); ansmax[i]:=askmax(1,l,r); end;end;procedure outit;var i:longint;begin for i:=1 to n-m do write(ansmin[i],' '); writeln(ansmin[n-m+1]); for i:=1 to n-m do write(ansmax[i],' '); writeln(ansmax[n-m+1]);end;begin assign(input,'window.in'); reset(input); assign(output,'window.out'); rewrite(output); init; buildtree(1,1,n); main; outit; close(input); close(output);end.