线段树区间更新 Count Color

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

这个的关键是用二进制存 每位的0，1表示这位有没有被涂过色 

每次输入的时候 就是1<<(add-1)  

所以更新到父节点的话 就要用 |

最后统计颜色 用与运算 都是1才为1 

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100005;
int sum[maxn*4];
struct node
{
    int l,r,val,mark;
}tree[maxn*4];
void build(int i,int l,int r)
{
    tree[i].mark=0;
    tree[i].val=1;
    tree[i].l=l,tree[i].r=r;
    //cout<<l<<" "<<r<<endl;
    if(l==r)return;


    int m=(l+r)/2;
    build(i*2,l,m);
    build(i*2+1,m+1,r);
}
void pushdown(int i)
{
     if(tree[i].mark!=0)
    {
    tree[i*2].mark=tree[i].mark;
    tree[i*2+1].mark=tree[i].mark;


    tree[i*2+1].val=tree[i].val;
    tree[i*2].val=tree[i].val;


    tree[i].mark=0;
    }
}
void update(int i,int l,int r,int add)
{


    //cout<<tree[i].l<<" "<<tree[i].r<<" "<<tree[i].mark<<" "<<tree[i].val<<endl;
    if(tree[i].l==l&&r==tree[i].r)    //要更新的区间比i所含的区间大 全部包含
    {
        tree[i].mark=1;
        tree[i].val=1<<(add-1);     //用不同位上的1表示
        //cout<<tree[i].l<<" "<<tree[i].r<<" "<<tree[i].mark<<" "<<tree[i].val<<endl;
        return;
    }
    if(tree[i].mark!=0)pushdown(i);


    int m=(tree[i].l+tree[i].r)/2;
    if(r<=m)update(i*2,l,r,add);
    else if(l>m)update(i*2+1,l,r,add);
    else
    {
        update(i*2,l,m,add);
        update(i*2+1,m+1,r,add);
    }
    tree[i].val=(tree[i*2].val)|(tree[i*2+1].val);
    //cout<<"CC"<<tree[i].l<<" "<<tree[i].r<<" "<<tree[i].mark<<" "<<tree[i].val<<endl;
    if(tree[i*2].mark&&tree[i*2+1].mark&&tree[i*2].val==tree[i*2+1].val)
    tree[i].mark=1;     //如果左右子树都被标记且颜色相同 就把父节点标记 这样查询的时候可以省时


}
int query(int i,int l,int r)
{
    //cout<<tree[i].l<<" "<<tree[i].r<<" "<<tree[i].mark<<" "<<tree[i].val<<endl;
    if(tree[i].l>=l&&tree[i].r<=r)
    return tree[i].val;
    pushdown(i);
    int ans=0;
    int m=(tree[i].l+tree[i].r)/2;
    if(l<=m)ans|=query(i*2,l,r);
    if(r>m)ans|=query(i*2+1,l,r);
    return ans;


}
int main()
{
    int l,t,o;
    while(scanf("%d%d%d",&l,&t,&o)!=-1)
    {
        char op;
        int a,b,c;
        memset(tree,0,sizeof(tree));
        build(1,1,l);
        while(o--)
        {
            scanf(" %c",&op);
            //cout<<op<<endl;
            if(op=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                if(a>b)swap(a,b);
                update(1,a,b,c);


            }
            else
            {
                scanf("%d%d",&a,&b);
                if(a>b)swap(a,b);
                int tmp=query(1,a,b);
                int count=0;
                for(int i=0;i<t;i++)
                {
                    if(tmp&(1<<i))count++;


                }
                cout<<count<<endl;
            }
        }


    }
    return 0;
}