A Corrupt Mayor's Performance Art (线段树+二进制压缩)

 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?

Input

There 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.

Output

For 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

题意：给出1-n个墙，初始的墙颜色都是2号颜色，然后两个操作，一个是给出一个区间[a,b]，和颜色c把这个区间颜色给改成c，另一个是给出区间[a,b]，把这个区间里包含的颜色打印出来

思路：思路很容易想到是线段树，但是怎么储存这个颜色是个问题，还得是按递增顺序的，一开始想用数组存，最后返回指针，但是最后写乱了，指针满天飞，就崩了，后来知道了用二级制压缩，什么意思呢，说白了就是用二进制表示形式00001010100101（举个栗子），来替代数组，到时候通过左移位<<或者右移位>>来实现c号颜色的更改

举个栗子：0    0   0  00000001010……1010001（二进制）

                  30 29 28……                       7654321（颜色的号码）

即从左往右数的

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1000100
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
typedef long long ll;
ll tre[maxn<<2],lazy[maxn<<2];//线段树的数组，懒惰标记数组
int n,m;
void Build(int l,int r,int rt){//建树 
	if(l==r){//到了叶子节点 
		tre[rt] = 2;//根据题意初始值为2 
		return;
	}
	int mid = l+r>>1;
	Build(lson);//左右进行递归建立左右子树 
	Build(rson);
	tre[rt] = tre[rt<<1] | tre[rt<<1|1];//返回给父亲两个儿子的信息，两个按位或就可以，这样两个儿子的信息就汇总在一起了 
}

void pushDown(int l,int r,int rt){//懒惰标记向下调整函数 
	lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];//把两个儿子懒惰标记上 
	tre[rt<<1] = lazy[rt];
	tre[rt<<1|1] = lazy[rt];//直接颜色覆盖，所以直接复制就可以 
	lazy[rt] = 0;//已经向下传递了，就可以取消父亲的懒惰标记了 
}

void Update(int a,int b,int c,int l,int r,int rt){//更新 
	if(a<=l&&r<=b){
		tre[rt] = 1<<(c-1);//更新的时候c号颜色，就是1从一号颜色左移c-1位到c号位置变成1 
		lazy[rt] = 1<<(c-1);//同理懒惰标记 
		return;
	}
	if(lazy[rt])pushDown(l,r,rt);//有懒惰标记向下传递 
	int mid = l+r>>1;
	if(a<=mid)Update(a,b,c,lson);//如果左子树有要查找的片段递归左子树 
	if(b>mid)Update(a,b,c,rson);//同理右边有要查找的找右子树 
	tre[rt] = tre[rt<<1] | tre[rt<<1|1];//修改完成汇总给父亲儿子信息 
}

int Query(int a,int b,int l,int r,int rt){
	if(a<=l&&r<=b){
		return tre[rt];//查询的时候找到相应区间直接返回值就可以 
	}
	if(lazy[rt])pushDown(l,r,rt);
	int mid = l+r>>1;
	ll ret = 0;
	if(a<=mid) ret |= Query(a,b,lson);//在查找返回的过程中直接汇总了信息|=等价于ret = ret | Query（）； 
	if(b>mid) ret |= Query(a,b,rson);
	return ret;
}
int main(){
	int i;
	while(~scanf("%d%d",&n,&m)){
		if(n==0&&m==0)break;
		memset(tre,0,sizeof(tre));
		memset(lazy,0,sizeof(lazy));//数组初始化
		Build(1,n,1);//建树
		while(m--){
			int a,b,c;
			char op[2];//这里用个数组便于输入 
			scanf("%s",op);
			if(op[0]=='P'){
				scanf("%d%d%d",&a,&b,&c);
				Update(a,b,c,1,n,1);
			}
			else{
				ll ans;
				scanf("%d%d",&a,&b);
				ans = Query(a,b,1,n,1);
				int flag=0;
				for(i = 1; i <= 30; i++){//返回一个值后，然后按位进行一位一位的比较是1的话看看在几号位置就是存在几号颜色 
					if(((ans>>(i-1))&1)&&flag==0){
						printf("%d",i);
						flag = 1;
					}
					else if((ans>>(i-1))&1){
					    printf(" %d",i);
					}
				}
				puts("");
			}
		}
	}
	return 0;
}