A Corrupt Mayor's Performance Art

Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money.

Because a lot of people praised mayor X's painting(of course, X was a mayor), mayor X believed more and more that he was a very talented painter. Soon mayor X was not satisfied with only making money. He wanted to be a famous painter. So he joined the local painting associates. Other painters had to elect him as the chairman of the associates. Then his painting sold at better price.

The local middle school from which mayor X graduated, wanted to beat mayor X's horse fart(In Chinese English, beating one's horse fart means flattering one hard). They built a wall, and invited mayor X to paint on it. Mayor X was very happy. But he really had no idea about what to paint because he could only paint very abstract paintings which nobody really understand. Mayor X's secretary suggested that he could make this thing not only a painting, but also a performance art work.

This was the secretary's idea:

The wall was divided into N segments and the width of each segment was one cun(cun is a Chinese length unit). All segments were numbered from 1 to N, from left to right. There were 30 kinds of colors mayor X could use to paint the wall. They named those colors as color 1, color 2 .... color 30. The wall's original color was color 2. Every time mayor X would paint some consecutive segments with a certain kind of color, and he did this for many times. Trying to make his performance art fancy, mayor X declared that at any moment, if someone asked how many kind of colors were there on any consecutive segments, he could give the number immediately without counting.

But mayor X didn't know how to give the right answer. Your friend, Mr. W was an secret officer of anti-corruption bureau, he helped mayor X on this problem and gained his trust. Do you know how Mr. Q did this?

InputThere are several test cases.

For each test case:

The first line contains two integers, N and M ,meaning that the wall is divided into N segments and there are M operations(0 < N <= 1,000,000; 0<M<=100,000)

Then M lines follow, each representing an operation. There are two kinds of operations, as described below:

1) P a b c
a, b and c are integers. This operation means that mayor X painted all segments from segment a to segment b with color c ( 0 < a<=b <= N, 0 < c <= 30).

2) Q a b
a and b are integers. This is a query operation. It means that someone asked that how many kinds of colors were there from segment a to segment b ( 0 < a<=b <= N).

Please note that the operations are given in time sequence.

The input ends with M = 0 and N = 0.OutputFor each query operation, print all kinds of color on the queried segments. For color 1, print 1, for color 2, print 2 ... etc. And this color sequence must be in ascending order.Sample Input

5 10
P 1 2 3
P 2 3 4
Q 2 3
Q 1 3
P 3 5 4
P 1 2 7
Q 1 3
Q 3 4
P 5 5 8
Q 1 5
0 0

Sample Output

4
3 4
4 7
4
4 7 8

線段樹 + 位運算，註釋寫的不少，由於不是很會線段樹，老是寫崩，邏輯要搞清楚。32位int正好能夠存30種顏色的狀態，存在第幾個顏色就把第幾位變爲1.

代碼：

///color n 用位移 1 << n來記錄
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#define lson l,mid,t << 1
#define rson mid + 1,r,t << 1 | 1
using namespace std;

int tree[4000005];
int lazy[4000005];///初始爲0 表示原來是什麼顏色 當更新的區間只佔子樹一部分時 拆分子樹向下更新lazy 無關的區間不受影響，只更新範圍內的區間

void build(int l,int r,int t)
{
    lazy[1] = 1 << 2;
    tree[t] = 1 << 2;///The wall's original color was color 2
    if(l == r)return;///線段樹的葉節點
    int mid = (l + r) >> 1;///從中間分紅左右子樹
    build(lson);
    build(rson);
}
void update(int L,int R,int col,int l,int r,int t)///更新
{
    if(r < L || l > R)return;///無交集
    if(l >= L && r <= R)///當前子樹處於查詢範圍內
    {
        lazy[t] = 1 << col;
        tree[t] = 1 << col;
        return;
    }
    if(lazy[t])///當前子樹的顏色都是lazy[t] 向下更新 若是值是0，表示左右子樹原本就不一致
    {
        lazy[t << 1] = lazy[t << 1 | 1] = lazy[t];
        tree[t << 1] = tree[t << 1 | 1] = tree[t];
        lazy[t] = 0;///當前子樹裏的顏色已經不是全都一致的了
    }
    int mid = (l + r) >> 1;
    update(L,R,col,lson);///當前子樹只有左子樹須要更新
    update(L,R,col,rson);///當前子樹只有右子樹須要更新

    tree[t] = tree[t << 1] | tree[t << 1 | 1];///向上更新  位或操做合併狀態
}
int query(int L,int R,int l,int r,int t)///查詢
{
    if(l > R || r < L)return 0;///再也不區間內返回0 表示沒顏色
    if(l >= L && r <= R || lazy[t])///子樹在查詢區間內 或者子樹狀態一致 直接返回
    {
        return tree[t];
    }
    int mid = (l + r) >> 1;
    return query(L,R,lson) | query(L,R,rson);///返回顏色狀態的並集
}
int main()
{
    int n,m,a,b,c;
    char ch[2];
    while(~scanf("%d%d",&n,&m)&&(n + m))
    {
        build(1,n,1);///建樹
        while(m --)
        {
            scanf("%s",ch);
            if(ch[0] == 'P')
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a,b,c,1,n,1);
            }
            else
            {
                scanf("%d%d",&a,&b);
                int ans = query(a,b,1,n,1),flag = 0;
                for(int i = 1;i <= 30;i ++)
                if((ans >> i) & 1)
                {
                    if(flag)printf(" %d",i);
                    else
                    {
                        flag = 1;
                        printf("%d",i);
                    }
                }
                putchar('\n');
            }
        }
    }
}