HDU5023:A Corrupt Mayor's Performance Art(线段树区域更新+二进制）

http://acm.hdu.edu.cn/showproblem.php?pid=5023

Problem 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

 

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

题意：一长由N小段组成的长条，每小段的初始颜色为2。现执行M个操作，每个操作是以下两种中的一种(0 < N <= 1,000,000; 0<M<=100,000) ：

P a b c ——> 将段a到段b涂成颜色c，c是1, 2, ... 30中的一种(0 < a<=b <= N, 0 < c <= 30)。

Q a b ——> 问段a到段b之间有哪几种颜色，按颜色大小从小到大输出(0 < a<=b <= N, 0 < c <= 30)。

——>>很明显此题可以用线段树实现Mlog(N)的解法。。

题目解析：这题收获很大，对二进制有了一点理解，一个int型数据占4个字节，总共32位，例如0的二进制为 00000000  00000000  00000000 00000000 ，而这题只有30种颜色，

所以完全可以用二进制来保存。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#define N 1000010
using namespace std;
struct node
{
    int l,r,lz,w;
} q[*N];
int n,m,res,tt;
void pushup(int rt)
{
    q[rt].w=q[rt<<].w|q[rt<<|].w;
}
void build(int l,int r,int rt)
{
    q[rt].l=l;
    q[rt].r=r;
    q[rt].lz=;
    q[rt].w=;
    if(l==r) return ;
    int mid=(l+r)>>;
    build(l,mid,rt<<);
    build(mid+,r,rt<<|);
    pushup(rt);
    return ;
}
void pushdown(int rt)
{
    if(q[rt].lz)
    {
        q[rt<<].lz=q[rt].lz;
        q[rt<<|].lz=q[rt].lz;
        q[rt<<].w=q[rt].lz;
        q[rt<<|].w=q[rt].lz;
        q[rt].lz=;
    }
}
void update(int lf,int rf,int l,int r,int rt,int key)
{
    if(lf<=l&&rf>=r)
    {
        q[rt].w=key;
        q[rt].lz=key;
        return ;
    }
    pushdown(rt);
    int mid=(l+r)>>;
    if(lf<=mid) update(lf,rf,l,mid,rt<<,key);
    if(rf>mid) update(lf,rf,mid+,r,rt<<|,key);
    pushup(rt);
    return ;
}
void query(int lf,int rf,int l,int r,int rt)
{
    if(lf<=l&&rf>=r)
    {
        res=res|q[rt].w;
        return ;
    }
    pushdown(rt);
    int mid=(l+r)>>;
    if(lf<=mid) query(lf,rf,l,mid,rt<<);
    if(rf>mid) query(lf,rf,mid+,r,rt<<|);
    return ;
}
int main()
{
    char ch[];
    int sum1,sum2,key;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==&&m==) break;
        build(,n,);
        for(int i=; i<m; i++)
        {
            scanf("%s",ch);
            if(ch[]=='P')
            {
                scanf("%d%d%d",&sum1,&sum2,&key);
                update(sum1,sum2,,n,,<<(key-));
            }
            else if(ch[]=='Q')
            {
                scanf("%d%d",&sum1,&sum2);
                res=;
                query(sum1,sum2,,n,);
                int flag=;
                for(int i=; i<; i++)
                {
                    if(res&(<<i)&&flag==)
                    {
                        printf(" %d",i+);
                    }
                    else if(res&(<<i)&&flag==)
                    {
                        printf("%d",i+);
                        flag=;
                    }
                }
                printf("\n");
            }
        }
    }
    return ;
}