Description
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.
题目大意：
#include<cstdio>
using namespace std;
const int MAXN=;
int tree[MAXN<<];
int lz[MAXN<<];
int n,t,o,L,R;
void up(int );
void down(int ,int ,int );
void col(int ,int ,int ,int );
int ask(int ,int ,int );
int ans(int );
int main()
{
int i,j,k;
char s[];
scanf("%d%d%d",&n,&t,&o);
lz[]=;
tree[]=;
for(i=;i<=o;i++){
scanf("%s",s);
if(s[]=='C'){
scanf("%d%d%d",&L,&R,&j);
if(L>=R)
k=L,L=R,R=k;
col(,n,,j);
}
else{
scanf("%d%d",&L,&R);
if(L>=R)
k=L,L=R,R=k;
j=ask(,n,);
printf("%d\n",ans(j));
}
}
}
void up(int nu){
tree[nu]=tree[nu<<]|tree[nu<<|];
}
void down(int l,int r,int nu){
if(!lz[nu])return ;
tree[nu<<]=lz[nu];
tree[nu<<|]=lz[nu];
lz[nu<<]=lz[nu<<|]=lz[nu];
lz[nu]=;
}
void col(int l,int r,int nu,int x){
if(L<=l&&r<=R){
tree[nu]=lz[nu]=(<<(x-));
return ;
}
down(l,r,nu);
int mid=(l+r)>>;
if(L<=mid)
col(l,mid,nu<<,x);
if(R>mid)
col(mid+,r,nu<<|,x);
up(nu);
}
int ask(int l,int r,int nu){
if(L<=l&&r<=R)
return tree[nu];
down(l,r,nu);
int ans=,mid=(l+r)>>;
if(L<=mid)
ans|=ask(l,mid,nu<<);
if(R>mid)
ans|=ask(mid+,r,nu<<|);
return ans;
}
int ans(int x){
int re=;
while(x){
if(x&)
re++;
x>>=;
}
return re;
}