POJ 2777 Count Color（线段树+位运算）

题目链接：http://poj.org/problem?id=2777

Description

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

Source

POJ Monthly--2006.03.26,dodo

题意：

给一个固定长度为L的画板

有两个操作：

C A B C：区间A--B内涂上颜色C。

P A B：查询区间AB内颜色种类数。

PS:

此题和HDU：5023是类似的！

附题解：http://blog.csdn.net/u012860063/article/details/39434665

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define LL int

const int maxn = 110017;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
    //把当前结点的信息更新到父结点
    sum[rt] = sum[rt<<1] | sum[rt<<1|1];//总共的颜色
}
void PushDown(int rt,int m)
{
    if(add[rt])
    {
        add[rt<<1] = add[rt];
        add[rt<<1|1] = add[rt];
        sum[rt<<1] = add[rt];
        sum[rt<<1|1] = add[rt];
        add[rt] = 0;//将标记向儿子节点移动后，父节点的延迟标记去掉
        //传递后，当前节点标记域清空
    }
}
void build(int l,int r,int rt)
{
    add[rt] = 0;//初始化为全部结点未被标记
    if (l == r)
    {
        sum[rt] = 1;//初始颜色为1
        return ;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        add[rt] =1<<(c-1);//位运算左移表示有某种颜色
        sum[rt] =1<<(c-1);
        return ;
    }
    PushDown(rt , r - l + 1);//----延迟标志域向下传递
    int mid = (l + r) >> 1;
    if (L <= mid)
        update(L , R , c , lson);//更新左儿子
    if (mid < R)
        update(L , R , c , rson);//更新右儿子
    PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        return sum[rt];
    }
    //要取rt子节点的值时，也要先把rt的延迟标记向下移动
    PushDown(rt , r - l + 1);
    int mid = (l + r) >> 1;
    LL ret = 0;
    if (L <= mid)
        ret |= query(L , R , lson);
    if (mid < R)
        ret |= query(L , R , rson);
    return ret;
}
int main()
{
    int L, T, O;
    int a, b, c;
    while(~scanf("%d%d%d",&L,&T,&O))
    {
        build(1, L, 1);//建树
        while(O--)//Q为询问次数
        {
            char op[2];
            scanf("%s",op);
            if(op[0] == 'P')
            {
                scanf("%d%d",&a,&b);
                if(a > b)
                {
                    int t = a;
                    a = b;
                    b = t;
                }
                LL tt=query(a, b, 1, L, 1);
                int ans = 0;
                while(tt)
                {
                    if(tt&1)
                    {
                        ans++;
                    }
                    tt>>=1;
                }
                printf("%d\n",ans);
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                if(a > b)
                {
                    int t = a;
                    a = b;
                    b = t;
                }
                update(a, b, c, 1, L, 1);
            }
        }
    }
    return 0;
}