poj2991 Crane（线段树）

Description

ACM has bought a new crane (crane -- jeřáb) . The crane consists of n segments of various lengths, connected by flexible joints. The end of the i-th segment is joined to the beginning of the i + 1-th one, for 1 ≤ i < n. The beginning of the first segment is fixed at point with coordinates (0, 0) and its end at point with coordinates (0, w), where w is the length of the first segment. All of the segments lie always in one plane, and the joints allow arbitrary rotation in that plane. After series of unpleasant accidents, it was decided that software that controls the crane must contain a piece of code that constantly checks the position of the end of crane, and stops the crane if a collision should happen.

Your task is to write a part of this software that determines the position of the end of the n-th segment after each command. The state of the crane is determined by the angles between consecutive segments. Initially, all of the angles are straight, i.e., 180^o. The operator issues commands that change the angle in exactly one joint.

Input

The input consists of several instances, separated by single empty lines.

The first line of each instance consists of two integers 1 ≤ n ≤10 000 and c 0 separated by a single space -- the number of segments of the crane and the number of commands. The second line consists of n integers l1,..., ln (1 li 100) separated by single spaces. The length of the i-th segment of the crane is li. The following c lines specify the commands of the operator. Each line describing the command consists of two integers s and a (1 ≤ s < n, 0 ≤ a ≤ 359) separated by a single space -- the order to change the angle between the s-th and the s + 1-th segment to a degrees (the angle is measured counterclockwise from the s-th to the s + 1-th segment).

Output

The output for each instance consists of c lines. The i-th of the lines consists of two rational numbers x and y separated by a single space -- the coordinates of the end of the n-th segment after the i-th command, rounded to two digits after the decimal point.

The outputs for each two consecutive instances must be separated by a single empty line.

Sample Input

Sample Output

5.00 10.00

-10.00 5.00

-5.00 10.00

Source

CTU Open 2005

题目大意：

题解：

我是没想到是线段树，看了是线段树后自己yy了一个建线段树的方法，好像跟书上的不一样。

书上的方法：

我的代码：

 program rrr(input,output);

 const

   eps=1e-10;

 type

   treetype=record

      l,r:longint;

      x,y,d:double;

   end;

 var

   a:array[..]of treetype;

   c:array[..]of double;

   b:array[..]of longint;

   n,m,i,x:longint;

   y,t,xx,yy:double;

 procedure build(k,l,r:longint);

 var

   mid,i:longint;

 begin

    a[k].l:=l;a[k].r:=r;a[k].x:=;a[k].d:=;

    if l=r then begin a[k].y:=b[l];exit; end;

    mid:=(l+r)>>;i:=k+k;

    build(i,l,mid);build(i+,mid+,r);

    a[k].y:=a[i].y+a[i+].y;

 end;

 procedure pushdown(k:longint);

 var

   i:longint;

 begin

    if a[k].l=a[k].r then a[k].d:=;

    if abs(a[k].d)<eps then exit;

    i:=k+k;

    xx:=a[i].x*cos(a[k].d)-a[i].y*sin(a[k].d);

    yy:=a[i].x*sin(a[k].d)+a[i].y*cos(a[k].d);

    a[i].x:=xx;a[i].y:=yy;a[i].d:=a[i].d+a[k].d;

    inc(i);

    xx:=a[i].x*cos(a[k].d)-a[i].y*sin(a[k].d);

    yy:=a[i].x*sin(a[k].d)+a[i].y*cos(a[k].d);

    a[i].x:=xx;a[i].y:=yy;a[i].d:=a[i].d+a[k].d;

    a[k].d:=;

 end;

 procedure change(k:longint);

 var

   mid,i:longint;

 begin

    pushdown(k);

    if x<a[k].l then

       begin

          xx:=a[k].x*cos(t)-a[k].y*sin(t);

          yy:=a[k].x*sin(t)+a[k].y*cos(t);

          a[k].x:=xx;a[k].y:=yy;

          a[k].d:=t;

          exit;

       end;

    mid:=(a[k].l+a[k].r)>>;i:=k+k;

    if x<mid then change(i);

    change(i+);

    a[k].x:=a[i].x+a[i+].x;a[k].y:=a[i].y+a[i+].y;

 end;

 begin

    assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);

    while not eof do

       begin

          readln(n,m);if (n=) and (m=) then break;

          for i:= to n do begin read(b[i]);c[i]:=pi; end;

          build(,,n);

          for i:= to m do

             begin

                readln(x,y);t:=y/*pi-c[x];c[x]:=y/*pi;

                change();

                writeln(a[].x::,' ',a[].y::);

             end;

          writeln;

       end;

    close(input);close(output);

 end.

巴特西

poj2991 Crane（线段树）

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