[MIPT 105]MRQ Promblem(线段树)

【题目大意】：

给你一组数，然后M个询问（x,y），询问第x+1到第y个当中的最小值

【分析】：

裸的线段树，也可以用传说中的ST算法~如果你不嫌麻烦也可以转化成LCA……

P.S. 效率方面，明显是静态的占优，所以写东西最好用数组……

【代码1(静态线段树)】：

program RMQ_MIPT; type node=record left,right:longint; min:real; end; var a:array[1..250000] of real; tree:array[1..1000000] of node; n,m,x,y,i:longint; function min2(x,y:real):real; begin if x<y then min2:=x else min2:=y; 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].min:=a[left]; exit; end; mid:=(left+right) div 2; buildtree(p*2,left,mid); buildtree(p*2+1,mid+1,right); tree[p].min:=min2(tree[p*2].min,tree[p*2+1].min); end; function findmin(p,left,right:longint):real; var mid:longint; begin if (tree[p].left=left) and (tree[p].right=right) then begin findmin:=tree[p].min; exit; end; mid:=(tree[p].left+tree[p].right) div 2; if right<=mid then begin findmin:=findmin(p*2,left,right); exit; end; if left>mid then begin findmin:=findmin(p*2+1,left,right); exit; end; findmin:=min2(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); for i:=1 to n do read(a[i]); buildtree(1,1,n); readln(m); for i:=1 to m do begin readln(x,y); x:=x+1; writeln(findmin(1,x,y):0:6); end; close(input); close(output); end.  

【代码2(动态线段树)】：

program RMQ; type link=^node; node=record left,right:longint; min:real; lc,rc:link; end; var root:link; a:array[1..250000] of real; m,n,i,x,y:longint; function min2(x,y:real):real; begin if x<y then min2:=x else min2:=y; end; procedure buildtree(var p:link;left,right:longint); var mid:longint; begin new(p); p^.left:=left; p^.right:=right; p^.lc:=nil; p^.rc:=nil; if left=right then begin p^.min:=a[left]; p^.lc:=nil; p^.rc:=nil; exit; end; mid:=(left+right) div 2; buildtree(p^.lc,left,mid); buildtree(p^.rc,mid+1,right); p^.min:=min2(p^.lc^.min,p^.rc^.min); end; function find(p:link;left,right:longint):real; var mid:longint; begin if (p^.left=left) and (p^.right=right) then begin find:=p^.min; exit; end; mid:=(p^.left+p^.right) div 2; if mid>=right then begin find:=find(p^.lc,left,right); exit; end; if mid<left then begin find:=find(p^.rc,left,right); exit; end; find:=min2(find(p^.lc,left,mid),find(p^.rc,mid+1,right)); end; begin assign(input,'a.in'); assign(output,'a.out'); reset(input); rewrite(output); readln(n); for i:=1 to n do read(a[i]); root:=nil; buildtree(root,1,n); readln(m); for i:=1 to m do begin readln(x,y); writeln(find(root,x+1,y):0:6); end; close(input); close(output); end.