HDU 5023 A Corrupt Mayor's Performance Art（线段树+优美的位运算）

题目链接：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
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：

P ：A到B区间图上第C种颜色；

Q：A到B所有点存在的颜色种类；

PS: 按照队友的思路代码讲解来得，给跪了；

思路：

就是利用优美的位运算的性质，初始为1，当有某种颜色时就<<代表此种颜色的数字，

最后在每次输出的时候再从1到30种颜色中逐一>>并&1就可以检验是否含有这种颜色！

PS:

此题和POJ：2777类似；

附题解：http://blog.csdn.net/u012860063/article/details/40433725

代码如下：

#include 
#include 
#include 
using namespace std;

#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define LL int

const int maxn = 1100017;
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt)
{
    //把当前结点的信息更新到父结点
    sum[rt] = sum[rt<<1] | sum[rt<<1|1];//总共的颜色
}
void PushDown(int rt,int m)
{
    if(add[rt])
    {
        add[rt<<1] = add[rt];
        add[rt<<1|1] = add[rt];
        sum[rt<<1] = add[rt] ;
        sum[rt<<1|1] = add[rt] ;
        add[rt] = 0;//将标记向儿子节点移动后，父节点的延迟标记去掉
        //传递后，当前节点标记域清空
    }
}
void build(int l,int r,int rt)
{
    add[rt] = 0;//初始化为所有结点未被标记
    if (l == r)
    {
        sum[rt]=2;//初始颜色为2
        return ;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        add[rt] =1<<(c-1);//位运算左移表示有某种颜色
        sum[rt] =1<<(c-1);
        return ;
    }
    PushDown(rt , r - l + 1);//----延迟标志域向下传递
    int mid = (l + r) >> 1;
    if (L <= mid)
        update(L , R , c , lson);//更新左儿子
    if (mid < R)
        update(L , R , c , rson);//更新右儿子
    PushUp(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
    if (L <= l && r <= R)
    {
        return sum[rt];
    }
    //要取rt子节点的值时，也要先把rt的延迟标记向下移动
    PushDown(rt , r - l + 1);
    int mid = (l + r) >> 1;
    LL ret = 0;
    if (L <= mid)
        ret |= query(L , R , lson);
    if (mid < R)
        ret |= query(L , R , rson);
    return ret;
}
int main()
{
    int N , Q;
    int a , b , c;
    while(scanf("%d%d",&N,&Q))
    {
        if(N==0 && Q==0)
            break;
        build(1 , N , 1);//建树
        while(Q--)//Q为询问次数
        {
            char op[2];
            scanf("%s",op);
            if(op[0] == 'Q')//Q为询问次数
            {
                scanf("%d%d",&a,&b);
                LL tt=query(a , b , 1 , N , 1);
                LL flag = 0;
                for(int i=1; i<=30; i++)
                {
                    if(tt>>(i-1)&1 && flag==0)
                    {
                        printf("%d",i);
                        flag = 1;
                    }
                    else if(tt>>(i-1)&1)
                        printf(" %d",i);
                }
                printf("\n");
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a , b , c , 1 , N , 1);
            }
        }
    }
    return 0;
}