HDU-A Corrupt Mayor's Performance Art【线段树+位运算】

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A Corrupt Mayor’s Performance Art

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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

题目思路：线段树，区间更新，30种颜色用二进制形式表示。

题目链接：HDU-5023

以下是代码：

#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;/*说明：    1. rt、L、R都是从1开始计数；    2. 线段树的空间应当是原数组的四倍；(最小应该为，向上取到最近的二的次幂，再乘以2)    3. 每次调用build，query和update时，L和R应该是整个区间（1-N），一般写成query(1,1,N)。 */ #define lson rt<<1,L,M#define rson rt<<1|1,M+1,R#define maxn 2000000 + 10int sumv[maxn<<2]; //树上结点int lazy[maxn<<2];//仅当区间更新时使用，点更新不需要 void pushUp(int rt) {    sumv[rt] = sumv[rt<<1] | sumv[rt<<1|1];}//区间更新时才需要使用pushDown void pushDown(int rt) {  //len在主函数里一般为　r - l +１     lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];  // lazy标记向两个儿子传递     sumv[rt<<1] = lazy[rt]; // 更新左儿子sumv     sumv[rt<<1|1] = lazy[rt];     // 更新右儿子sumv     lazy[rt] = 0;                            // 向下传递完后清空当前结点lazy标记 }void build(int rt, int L, int R) {    int M = (L+R) >> 1;    if (L == R) {        sumv[rt] = 2; //始化线段树。    } else {        build(lson);        build(rson);        pushUp(rt);    }}//区间更新 void update(int tl, int tr, int v, int rt, int L, int R){    int M = (L+R)>>1;    if (tl<=L && R<=tr) {        lazy[rt] = v;        sumv[rt] = v;    } else {        if (lazy[rt]) pushDown(rt, R-L+1);        if (tl <= M) update(tl, tr, v, lson);        if (tr >= M+1) update(tl, tr, v, rson);        pushUp(rt);    }}int query(int tl, int tr, int rt, int L, int R) {    int M = (L+R) >> 1;    int ret = 0;     if (lazy[rt]) pushDown(rt, R-L+1);    if (tl<=L && R<=tr) return sumv[rt]; // 查询区间涵盖了整个区间     if (tl <= M) ret |= query(tl, tr, lson); // 查询区间在左边有分布     if (M < tr) ret |= query(tl, tr, rson); // 查询区间在右边有分布    return ret; }int main(){    int n,m;    while(scanf("%d%d",&n,&m) != EOF)    {        memset(lazy,0,sizeof(lazy));        if (n == 0 && m == 0) break;        build(1,1,n);        while(m--)        {            char ch;            scanf("%*c%c",&ch);                     if (ch == 'P')            {                int x,y,value;                scanf("%d%d%d",&x,&y,&value);                update(x,y,1 << (value - 1),1,1,n);             }            else            {                int x,y;                scanf("%d%d",&x,&y);                int ans = query(x,y,1,1,n);                int first = 1;                for (int i = 1; i <= 32; i++)                {                    if (ans & 1)                    {                        if (first)                        {                            first = 0;                            printf("%d",i);                        }                        else                        {                            printf(" %d",i);                        }                    }                    ans >>= 1;                }                printf("\n");            }        }    }    return 0;}