poj 2777 ——线段树基础
Count Color
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16872 | Accepted: 4839 |
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.
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
这里依然采用lazy思想（参看上一篇）。
type
ji=^rec;
rec=record
color,l,r:longint;
lson,rson:ji;
end;
var
v:array[-1..30] of boolean;
ans,i,j,k,m,n,t,s,sum:longint;
a:ji;
ch:char;
procedure build(var a:ji; l,r:longint);
var
mid:longint;
begin
new(a); a^.l:=l; a^.r:=r;
if r>l+1 then
begin
mid:=(l+r)>>1;
build(a^.lson,l,mid);
build(a^.rson,mid,r);
end
else
begin
a^.lson:=nil;
a^.rson:=nil;
end;
end;
procedure down(var a:ji);
begin
if a^.color>0 then
begin
if a^.lson<>nil then
a^.lson^.color:=a^.color;
if a^.rson<>nil then
a^.rson^.color:=a^.color;
a^.color:=-1;
end;
end;
procedure insert(var a:ji; l,r,s:longint);
var
mid:longint;
begin
if (l<=a^.l)and(a^.r<=r) then a^.color:=s else
begin
down(a);
mid:=(a^.l+a^.r)>>1;
if l<mid then insert(a^.lson,l,r,s);
if r>mid then insert(a^.rson,l,r,s);
end;
end;
procedure count(var a:ji; l,r:longint);
var
mid:longint;
begin
v[a^.color]:=true;
if a^.color<0 then
begin
mid:=(a^.l+a^.r)>>1;
if l<mid then count(a^.lson,l,r);
if r>mid then count(a^.rson,l,r);
end;
end;
begin
readln(n,sum,m);
build(a,0,n);
a^.color:=1;
for i:=1 to m do
begin
read(ch,s,t);
if ch='C' then
begin
read(k);
insert(a,s-1,t,k);
end
else
begin
ans:=0;
fillchar(v,sizeof(v),false);
count(a,s-1,t);
for j:=1 to sum do
if v[j] then inc(ans);
writeln(ans);
end;
readln;
end;
end.