Count Color POJ - 2777 线段树

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
线段树
用一个LL 位表示颜色情况
用一个laz表示当前块是否是同一颜色的
pushdown
如果当前颜色是统一的，那么将这个颜色推给下面，同时将下面的laz标识设置为1， 然后清除该标识

pushup
color是左右子节点color的按位与
如果 左边颜色是统一的！ 而且 右边颜色是统一的！ 而且左右两边颜色相同！
才讲该节点的laz设置为1
在查询的时候由于我们查询的是color，如果当前区间的laz = 1
表示区间内所有颜色都是统一的，那么我们可以直接返回当前节点的颜色

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
#define MAXN 100009
LL l,t,o;
struct node
{
    LL l,r;
    LL laz;
    LL sum;//用一个数字的每个位表示这个段有多少颜色
}T[MAXN *  + ];
void pushdown(LL p)
{
    if(T[p].laz)
    {
        T[p].laz = ;
        T[p*].laz = T[p*+].laz = ;
        T[p*].sum = T[p* + ].sum = T[p].sum;
    }
}
void pushup(LL p)
{
    T[p].sum = T[p*].sum | T[p*+].sum;
    if(T[p*].laz && T[p*+].laz && T[p*].sum == T[p* + ].sum)
        T[p].laz = ;
}
void build(LL x,LL l,LL r)
{
    T[x].l = l,T[x].r = r;
    if(l == r)
    {
        return ;
    }
    LL mid = (l + r)/;
    build(x * ,l ,mid);
    build(x *  + ,mid + ,r);
    //pushup(x);
}
void update(LL x,LL l,LL r,LL val)
{
    if(T[x].l == l&&T[x].r == r)
    {
        T[x].sum = ( << (val - ) );
        T[x].laz = ;
        return ;
    }
    pushdown(x);
    LL mid = ( T[x].l + T[x].r)/;
    if(r<=mid)
        update(x * , l ,r, val);
    else if(l > mid)
        update(x * +, l , r, val);
    else
    {
        update(x *  ,l, mid,val);
        update(x *  + ,mid + , r, val);
    }
    pushup(x);
}
LL query(LL x, LL l, LL r)
{
    if(T[x].laz || (T[x].l == l && T[x].r == r))
    {
        return T[x].sum;
    }
//    pushdown(x);
    LL  mid = (T[x].l + T[x].r )/;
    if(r<=mid)
        return query(x * , l, r);
    else if( l > mid)
        return query(x *  + , l ,r);
    else
    {
    //    LL tmp1 = query(x *2, l, mid);
    //    LL tmp2 = query(x*2 +1, mid + 1, r);
        return query(x *, l, mid) | query(x* +, mid + , r);
    }
}
int main()
{

    char c[];
    LL L,R,tmp;
    scanf("%lld%lld%lld",&l,&t,&o);
        build(,,l);
        T[].laz = T[].sum = ;
        for(LL i = ;i<o;i++)
        {
            scanf("%s%lld%lld",c,&L,&R);
            if(L > R)
            {
                LL sd = L;
                L = R;
                R = sd;
            }
            if(c[]=='C')
            {
                scanf("%lld",&tmp);
                update(,L,R,tmp);
            }
            else if( c[] == 'P')
            {
                LL tmp = query(,L,R);
                LL cnt = ;
                for(int i = ; i < t; i++)
                {
                    if(tmp&(<<i))
                        cnt++;
                }
                printf("%lld\n",cnt);
            }
        }

}