POJ 2155(二维zkw线段树）-CFANZ编程社区

Default Matrix Description 给定一个矩阵（初始为0），维护2个操作： 1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) ,以(x1,y1)为左上角，(x2,y2)为右上角的矩阵取反。 2. Q x y (1 <= x, y <= n) 输出(x,y)的状态 Input 第一行为数据组数X (X <= 10) 每组数据第一行为N,T (2 <= N <= 1000, 1 <= T <= 50000) 表示矩阵边长与操作次数。接下来T行，为Q或C操作 Output 请输出所有提问结果。每组数据后一回车。 Sample Input 12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1 Sample Output 1001 Source POJ Monthly,Lou Tiancheng

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 13137		Accepted: 4948

这题是二维的线段树，

先建一棵线段树，再将它的所有结点（包括根）建立一个二叉树（维护这个结点）

异或仅需记录标记某点所对应的区间取反

查找时统计共记了几个标记即可（标记永久化）

Program Matrix;
const
   maxtt=10;
   maxn=1000;
   maxMM=50000;
var
   tt,n,mm,i,j,k,x1,y1,x2,y2,x,y,p1,p2,ans:longint;
   c:char;
   t:array[1..5000,1..5000] of boolean;
   M:longint;

Procedure change_y(x,y1,y2:longint);
var
   i,j:longint;
begin
   dec(y1);inc(y2);
   inc(y1,M);inc(y2,M);
   while ((y1 xor y2 xor 1)>0) do
   begin
      if ((y1 and 1)=0) then t[x,y1+1]:=not(t[x,y1+1]);
      if ((y2 and 1)=1) then t[x,y2-1]:=not(t[x,y2-1]);
      y1:=y1 shr 1;y2:=y2 shr 1;
   end;

end;

Procedure change_x(x1,y1,x2,y2:longint);
var
   i,j:longint;
begin
   dec(x1);inc(x2);
   inc(x1,M);inc(x2,M);
   while ((x1 xor x2 xor 1)>0) do
   begin
      if ((x1 and 1)=0) then change_y(x1+1,y1,y2);
      if ((x2 and 1)=1) then change_y(x2-1,y1,y2);
      x1:=x1 shr 1;x2:=x2 shr 1;
   end;
end;

function find_y(x,y:longint):boolean;
var
   i,j:longint;
begin
   inc(y,M); find_y:=false;
   while (y>0) do
   begin
      if (t[x,y]) then find_y:=not(find_y);
      y:=y shr 1;
   end;

end;
function find_x(x,y:longint):boolean;
var
   i,j:longint;
begin
   inc(x,M); find_x:=false;
   while (x>0) do
   begin
      find_x:=find_x xor find_y(x,y);
      x:=x shr 1;
   end;
end;
begin
   readln(tt);
   while (tt>0) do
   begin
      fillchar(t,sizeof(t),false);
      readln(n,mm);
      M:=1;
      while (M-2<n) do M:=M shl 1;
      for k:=1 to mm do
      begin
         read(c);
         if c='C' then
         begin
            readln(x1,y1,x2,y2);
            change_x(x1,y1,x2,y2);





         end
         else
         begin     //Q
            readln(x1,y1);
            if (find_x(x1,y1)) then writeln('1') else writeln('0');


         end;



      end;
      dec(tt);
      if (tt>0) then writeln;
   end;
end.