Count Color（线段树＆状压）

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.

染色问题，30个颜色，所以可以用一个int类的数字表示所有颜色。那么就可以把所有的颜色状态都表示出来了。但是要注意输入的左右值不一定那个大，所以要是l>r的时候要交换一下。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e5+5;
struct node
{
    int l, r;
    int color, lazy;
}tree[maxn<<2];

inline void PushUp(int cur)
{
    tree[cur].color = tree[cur<<1].color|tree[cur<<1|1].color;
}

inline void PushDown(int cur)
{
    if(tree[cur].lazy == 0) return ;
    tree[cur<<1].lazy = tree[cur<<1|1].lazy = tree[cur].lazy;
    tree[cur<<1].color = tree[cur<<1|1].color = tree[cur].lazy;
    tree[cur].lazy = 0;
}

void build(int l, int r, int cur)
{
    tree[cur].l = l, tree[cur].r = r;   tree[cur].lazy = 0;
    if(l == r)
    {
        tree[cur].color = (1<<1);
        return ;
    }
    int mid = (l+r)>>1;
    build(l, mid, cur<<1);
    build(mid+1, r, cur<<1|1);
    PushUp(cur);
}

void update(int l, int r, int cur, int val)
{
    if(l <= tree[cur].l && tree[cur].r <= r)
    {
        tree[cur].lazy = tree[cur].color = (1<<val);
        return ;
    }
    PushDown(cur);
    int mid = (tree[cur].l+tree[cur].r)>>1;
    if(l <= mid)    update(l, r, cur<<1, val);
    if(r > mid) update(l, r, cur<<1|1, val);
    PushUp(cur);
}

int query(int l, int r, int cur)
{
    if(l <= tree[cur].l && tree[cur].r <= r)
        return tree[cur].color;
    PushDown(cur);
    int mid = (tree[cur].l+tree[cur].r)>>1;
    int result = 0;
    if(l <= mid)    result |= query(l, r, cur<<1);
    if(r > mid) result |= query(l, r, cur<<1|1);
    return result;
}

int main()
{
    int n, color_num, op_num;
    char op[5];
    cin >> n >> color_num >> op_num;
    build(1, n, 1);
    for(int i = 1; i <= op_num; i++)
    {
        int l, r, val;
        scanf("%s", op);
        if(op[0] == 'C')
        {
            scanf("%d%d%d", &l, &r, &val);
            if(l > r)   swap(l, r);
            update(l, r, 1, val);
        }
        else
        {
            scanf("%d%d", &l, &r);
            if(l > r)   swap(l, r);
            int ans = 0, temp = query(l, r, 1);
            while(temp)
            {
                if(temp&1)  ans++;
                temp >>= 1;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}