[PKU 2823]Sliding Windows(线段树版)

【题目大意】：

有N个数，每次从左至右选M个数，即[1..M]，[2..M+1]，...，[M-N+1..N]，选取每个区间的最大值和最小值。

 

【题目分析】：

这个很明显是维护区间性质，所以最容易想到线段树。可以作为线段树的基础练习题来练习~

 

【代码】：

program PKU_2823;type node=record left,right,max,min:longint; end;var ans1,ans2,a:array[1..1000000] of longint; tree:array[1..3000000] of node; n,m,i:longint;function max(a,b:longint):longint;begin if a>b then exit(a) else exit(b);end;function min(a,b:longint):longint;begin if a<b then exit(a) else exit(b);end;procedure buildtree(p,left,right:longint);var mid:longint;begin tree[p].left:=left; tree[p].right:=right; if left=right then begin tree[p].max:=a[left]; tree[p].min:=a[left]; exit; end; mid:=(left+right) shr 1; buildtree(p*2,left,mid); buildtree(p*2+1,mid+1,right); tree[p].max:=max(tree[p*2].max,tree[p*2+1].max); tree[p].min:=min(tree[p*2].min,tree[p*2+1].min);end;function findmax(p,left,right:longint):longint;var mid:longint;begin if (tree[p].left=left) and (tree[p].right=right) then exit(tree[p].max); mid:=(tree[p].left+tree[p].right) shr 1; if right<=mid then exit(findmax(p*2,left,right)) else if left>mid then exit(findmax(p*2+1,left,right)) else exit(max(findmax(p*2,left,mid),findmax(p*2+1,mid+1,right)));end;function findmin(p,left,right:longint):longint;var mid:longint;begin if (tree[p].left=left) and (tree[p].right=right) then exit(tree[p].min); mid:=(tree[p].left+tree[p].right) shr 1; if right<=mid then exit(findmin(p*2,left,right)) else if left>mid then exit(findmin(p*2+1,left,right)) else exit(min(findmin(p*2,left,mid),findmin(p*2+1,mid+1,right)));end;begin assign(input,'a.in'); assign(output,'a.out'); reset(input); rewrite(output); readln(n,m); for i:=1 to n do read(a[i]); buildtree(1,1,n); for i:=1 to n-m+1 do begin ans1[i]:=findmin(1,i,i+m-1); ans2[i]:=findmax(1,i,i+m-1); end; write(ans1[1]); for i:=2 to n-m+1 do write(' ',ans1[i]); writeln; write(ans2[1]); for i:=2 to n-m+1 do write(' ',ans2[i]); writeln; close(input); close(output);end.