【HDU 5023】A Corrupt Mayor's Performance Art 【线段树+状态压缩】

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： n个点起始全部有相同的一种颜色， 然后有m次操作，P a b c 每次将区间[a,b]都染色为c，Q a b 将区间[a,b]之间的所有不同的颜色按照升序输出。

分析: 才30种颜色，我们可以用int型来状态压缩，一个二进制位代表一个颜色，然后用线段树来维护区间修改和查询，区间颜色的增加用 | 运算符就可以来维护。对于每次查询，得到最终的颜色状态，然后遍历一遍找到所有颜色。

代码

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = +;
const int MAXM = ;
const int mod = +;
const int inf = ;

struct Tree{
    int l,r;
    int col;
    int lazy;
}tree[MAXN<<];

void Up(int o) { tree[o].col=tree[o<<].col | tree[o<<|].col; }
void Down(int o){
    if(tree[o].lazy!=-){
        tree[o<<].lazy=tree[o<<|].lazy=tree[o].lazy;
        tree[o<<].col=tree[o<<|].col=(<<tree[o].lazy);
        tree[o].lazy=-; 
    }
}
void Build(int o,int le,int ri){
    tree[o].l=le;tree[o].r=ri;tree[o].lazy=-;
    tree[o].col=;
    if(le==ri) return ;
    int mid=(le+ri)>>;
    Build(o<<,le,mid);
    Build(o<<|,mid+,ri);
    Up(o);
}
void Update(int o,int le,int ri,int val){
    if(tree[o].l>=le&&tree[o].r<=ri) {
        tree[o].lazy=val;
        tree[o].col=(<<val);
        return ;
    }
    Down(o);
    int mid=(tree[o].l+tree[o].r)>>;
    if(ri<= mid ) Update(o<<,le,ri,val);
    else if(le>mid) Update(o<<|,le,ri,val);
    else  {
        Update(o<<,le,mid,val);
        Update(o<<|,mid+,ri,val);
    }
    Up(o);
}
int Query(int o,int le,int ri){
    if(tree[o].l>=le&&tree[o].r<=ri) return tree[o].col;
    Down(o);
    int mid=(tree[o].l+tree[o].r)>>;
    if(ri<=mid) return Query(o<<,le,ri);
    else if(le>mid) return Query(o<<|,le,ri);
    else  return Query(o<<,le,mid)|Query(o<<|,mid+,ri);
}

int main(){
    CLOSE();
//  fread();
//  fwrite();
    int n,m;
    while(scanf("%d%d",&n,&m)&&(n||m)){
        Build(,,n);
        char op[];int a,b,c;
        while(m--){
            scanf("%s%d%d",op,&a,&b);
            if(op[]=='P') {
                scanf("%d",&c);
                Update(,a,b,c-);
            }else {
                int ans=Query(,a,b);
                vector<int>ve;
                for(int i=;i<+;i++)
                    if(ans&(<<i))  ve.push_back(i+);
                for(int i=;i<ve.size();i++) 
                    printf("%d%s",ve[i],i!=ve.size()-?" ":"");
                if(ve.size()) puts("");
            }
        }
    }   
    return ;
}