poj2777 Count Color

Description

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. 

There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment
with only one color. We can do following two operations on the board: 

1. "C A B C" Color the board from segment A to segment B with color C. 

2. "P A B" Output the number of different colors painted between segment A and segment B (including). 

In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the
beginning, the board was painted in color 1. Now the rest of problem is left to your. 

Input

First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation
defined previously.
Output
Ouput results of the output operation in order, each line contains a number.
题解：
令线段树节点定义为: l,r,value,flag分被表示区间左右端点，该区间的涂色为value,flag为标记量，true表示该区间涂色为最新的涂色，否则表示该区间的子区间有更新的涂色。由于初始时均涂色为1，故value的值均为1，flag均为true(即都是最新的涂色)。本题的难点在于更新和查找。

代码：
with tree[p] do  

    begin  

      if (l=b) and (r=e) then  

        begin  

          color:=1 shl (c-1);  

          cover:=true;  

          exit;  

        end;  

      if cover then  

        begin  

          cover:=false;  

          tree[p*2].cover:=true;  

          tree[p*2].color:=color;  

          tree[p*2+1].cover:=true;  

          tree[p*2+1].color:=color;  

        end;  

      m:=(l+r) div 2;  

      if e<=m then ins(p*2,b,e,c) else  

        if b>m then ins(p*2+1,b,e,c) else  

          begin  

            ins(p*2,b,m,c);  

            ins(p*2+1,m+1,e,c);  

          end;  

      color:=tree[p*2].color or tree[p*2+1].color;  

    end;