JZOJ 4603 颜料大乱斗【NOIP2016提高A组模拟7.15】

颜色大乱斗

题目描述

输入格式

输出格式

样例输入

2 2 4

C 1 1 2

P 1 2

C 2 2 2

P 1 2

样例输出

2

1

数据范围

题解

这一题，题目要求维护加询问区间，那我们种30棵线段树维护一段区间中每种颜色的种类数即可。

记得在打线段树时加上懒惰迁移。

统计答案和更新区间以及懒惰标记下传时，只需用c的时间去做就可以了，反正时间复杂度不会太大。

Code(Pascal)

const
kt=1;
var
ch:char;
tj:array[0..100] of longint;
n,c,m,i,j,k,l,o,p,ans,x,y,t,u:longint;
la:array[0..300000] of longint;
tr:array[0..300000,0..30] of longint;
procedure jl(o,l,r:longint);
var
mid:longint;
begin
tr[o,kt]:=r-l+1;
if l=r then exit;
mid:=(l+r) div 2;
jl(o*2,l,mid);
jl(o*2+1,mid+1,r);
end;
procedure draw(o,l,r,ll,rr,cqy:longint);
var
ls,rs,i,mid:longint;
begin
mid:=(l+r) div 2;
ls:=O*2;
rs:=ls+1;
if (l=ll) and (r=rr) then
begin
la[o]:=cqy;
for i:=kt to c do
tr[o,i]:=0;
tr[o,cqy]:=r-l+1;
exit;
end;
if la[o]>0 then
begin
for i:=kt to c do
begin
tr[ls,i]:=0;
tr[rs,i]:=0;
end;
tr[ls,la[o]]:=mid-l+1;
tr[rs,la[o]]:=r-mid;
la[ls]:=la[o];
la[rs]:=la[o];
la[o]:=0;
end;
if rr<=mid then draw(ls,l,mid,ll,rr,cqy)
else if ll>mid then draw(rs,mid+1,r,ll,rr,cqy)
else
begin
draw(ls,l,mid,ll,mid,cqy);
draw(rs,mid+1,r,mid+1,rr,cqy);
end;
for i:=kt to c do
tr[o,i]:=tr[ls,i]+tr[rs,i];
end;
procedure cx(o,l,r,ll,rr:longint);
var
ls,rs,mid,i:longint;
begin
if (l=ll) and (r=rr) then
begin
for i:=kt to c do
tj[i]:=tj[i]+tr[o,i];
exit;
end;
mid:=(l+r) div 2;
ls:=o*2;
rs:=ls+1;
if la[o]>0 then
begin
for i:=kt to c do
begin
tr[ls,i]:=0;
tr[rs,i]:=0;
end;
tr[ls,la[o]]:=mid-l+1;
tr[rs,la[o]]:=r-mid;
la[ls]:=la[o];
la[rs]:=la[o];
la[o]:=0;
end;
if rr<=mid then cx(ls,l,mid,ll,rr)
else if ll>mid then cx(rs,mid+1,r,ll,rr)
else
begin
cx(ls,l,mid,ll,mid);
cx(rs,mid+1,r,mid+1,rr);
end;
for i:=kt to c do
tr[o,i]:=tr[ls,i]+tr[rs,i];
end;
begin
readln(n,c,m);
jl(1,1,n);
for i:=1 to m do
begin
read(ch);
if ch='C' then
begin
readln(x,y,t);
if x>y then
begin
u:=x;
x:=y;
y:=u;
end;
draw(1,1,n,x,y,t);
end
else
begin
readln(x,y);
if x>y then
begin
u:=x;
x:=y;
y:=u;
end;
for l:=kt to c do
tj[l]:=0;
cx(1,1,n,x,y);
ans:=0;
for l:=kt to c do
if tj[l]>0 then inc(ans);
writeln(ans);
end;
end;
end.