hdu--5023 A Corrupt Mayor's Performance Art（线段树+区间更新+位运算）

Description

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：给出点[1,n]和更新查询次数m，P是对区间[a,b]上的点染色成c，Q是查询区间[a,b]中有哪些颜色，从小到大输出。
思路：建树，用ins对区间进行更新并进行懒惰标记，然后用sea进行查询。由于颜色有30种，所以可以用int的二进制进行储存。
注意：当建树较小时会TLE而不是MLE或RE，让我纠结了好久
AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#define ll long long
using namespace std;
const int maxn=1000000+10;
struct note
{
    int l,r,ans;
    bool t;
} a[maxn<<2];//注意：maxn<<2
int build(int l,int r,int k)
{
    a[k].l=l,a[k].r=r,a[k].ans=1<<2,a[k].t=0;
    if(l==r)
    {
        return 0;
    }
    int mid=(l+r)>>1;
    build(l,mid,k<<1);
    build(mid+1,r,k<<1|1);
    return 0;
}
int pushdown(int k)
{
        a[k<<1].ans=a[k<<1|1].ans=a[k].ans;
        a[k].t=0;
        a[k<<1].t=a[k<<1|1].t=1;
    return 0;
}
int ins(int l,int r,int k,int d)
{
    if(a[k].l==l&&a[k].r==r)
    {
        a[k].t=1;
        a[k].ans=1<<d;
        return 0;
    }
    if(a[k].t)   pushdown(k);
    int mid=(a[k].l+a[k].r)>>1;
    if(a[k<<1].r>=r)  ins(l,r,k<<1,d);
    else if(a[k<<1|1].l<=l) ins(l,r,k<<1|1,d);
    else
    {
        ins(l,mid,k<<1,d);
        ins(mid+1,r,k<<1|1,d);
    }
    a[k].ans=a[k<<1].ans|a[k<<1|1].ans;
    return 0;
}
int sea(int l,int r,int k)
{
    if(l==a[k].l&&r==a[k].r)
    {
        return a[k].ans;
    }
    if(a[k].t) pushdown(k);
    int mid=(a[k].l+a[k].r)>>1,ans;
    if(a[k<<1].r>=r) ans=sea(l,r,k<<1);
    else if(a[k<<1|1].l<=l) ans=sea(l,r,k<<1|1);
    else
    {
        ans=sea(l,mid,k<<1)|sea(mid+1,r,k<<1|1);
    }
    return ans;
}
int main()
{
    int n,m,a,b,c;
    char  s[10];
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        build(1,n,1);
        while(m--)
        {
            scanf("%s",s);
            if(s[0]=='P')
            {
                scanf("%d%d%d",&a,&b,&c);
                ins(a,b,1,c);
            }
            else
            {
                int ans,i;
                scanf("%d%d",&a,&b);
                ans=sea(a,b,1);
                for(i=1; i<=30; i++)//注意：i<=30
                    if((ans>>i)&1)
                    {
                        printf("%d",i);
                        break;
                    }
                for(i++;i<=30;i++)
                    if((ans>>i)&1)
                    printf(" %d",i);
                printf("
");
            }
        }
    }
    return 0;
}