poj2777 Count Color 2011-12-20

Count Color

Time Limit: 1000MSMemory Limit: 65536K

Total Submissions: 23937Accepted: 7078

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.

Sample Input

2 2 4

C 1 1 2

P 1 2

C 2 2 2

P 1 2

Sample Output

2

1

Source

POJ Monthly--2006.03.26,dodo

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题目大意：给一个初始为颜色1的板染色，询问一个区间有多少种颜色。

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颜色最多只有30种，所以需要用一个30位的二进制数表示每段的状态，用线段树修改和查询，基础的线段树水题。= =

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  1 Program Stone;
  2
  3 var n,o,m,lc,rc,x,ans:longint;
  4
  5     a:array[1..1 shl 18]of longint;
  6
  7     b:array[1..1 shl 18]of boolean;
  8
  9
 10
 11  procedure update(head,tail,num:longint);    //修改
 12
 13  var i,j,k:longint;
 14
 15   begin
 16
 17    if (head>=lc)and(tail<=rc) then          //如果目前访问的区间被所需修改的区间包括，则整个区间标为x色。x为当前要涂成的颜色
 18
 19      begin
 20
 21       a[num]:=1 shl (x-1);
 22
 23       b[num]:=true;                        //b表示该区间是否都为同一种颜色。
 24
 25       exit;                                //因为整个区间修改，所以可以直接返回。
 26
 27      end;
 28
 29    if b[num] then begin
 30
 31                    a[num*2]:=a[num];a[num*2+1]:=a[num];       //将该区间的状态传递给左右孩子区间。
 32
 33                    b[num*2]:=true;b[num*2+1]:=true;
 34
 35                    b[num]:=false;                             //并修改标记。
 36
 37                   end;
 38
 39    k:=(head+tail)div 2;
 40
 41    if k>=rc then update(head,k,num*2) else
 42
 43    if lc>k  then update(k+1,tail,num*2+1) else
 44
 45     begin
 46
 47       update(head,k,num*2);
 48
 49       update(k+1,tail,num*2+1);
 50
 51     end;                                                    //分别访问孩子区间。
 52
 53    a[num]:=a[num*2]or a[num*2+1];                           //使用or计算当前区间的颜色数。
 54
 55   end;
 56
 57
 58
 59  procedure query(head,tail,num:longint);        //询问
 60
 61  var i,j,k:longint;
 62
 63   begin
 64
 65    if (head>=lc)and(tail<=rc) then
 66
 67      begin
 68
 69       ans:=ans or a[num];                      //计算答案
 70
 71       exit;
 72
 73      end;
 74
 75    if b[num] then begin
 76
 77                    a[num*2]:=a[num];a[num*2+1]:=a[num];
 78
 79                    b[num*2]:=true;b[num*2+1]:=true;
 80
 81                    b[num]:=false;
 82
 83                   end;
 84
 85    k:=(head+tail)div 2;
 86
 87    if k>=rc then query(head,k,num*2) else
 88
 89    if lc>k  then query(k+1,tail,num*2+1) else
 90
 91     begin
 92
 93       query(head,k,num*2);
 94
 95       query(k+1,tail,num*2+1);
 96
 97     end;
 98
 99   end;
100
101
102
103  procedure init;
104
105  var i,j,k:longint;
106
107      c,sp:char;
108
109   begin
110
111    fillchar(b,sizeof(b),false);
112
113    a[1]:=1;b[1]:=true;               //初始化，将整块板标为颜色1。
114
115    readln(n,o,m);
116
117     for i:=1 to m do
118
119      begin
120
121       read(c,sp);
122
123       if c=‘C‘ then begin
124
125                       readln(lc,rc,x);
126
127                       if rc<lc then begin
128
129                                      k:=rc;rc:=lc;lc:=k;
130
131                                     end;
132
133                       update(1,n,1);
134
135                     end
136
137                else begin
138
139                       readln(lc,rc);
140
141                       if rc<lc then begin
142
143                                      k:=rc;rc:=lc;lc:=k;
144
145                                     end;
146
147                       ans:=0;
148
149                       query(1,n,1);
150
151                       k:=0;
152
153                       for j:=0 to o-1 do
154
155                        if (ans shr j) and 1=1 then inc(k);   //查看二进制的数有多少位为1，即总共有多少种颜色。
156
157                       writeln(k);
158
159                     end;
160
161      end;
162
163   end;
164
165 Begin
166
167  assign(input,‘poj2777.in‘);reset(input);
168
169   init;
170
171  close(input);
172
173 end.