A Corrupt Mayor's Performance Art(线段树区间更新+位运算,颜色段种类)
A Corrupt Mayor's Performance Art
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others) Total Submission(s): 1905 Accepted Submission(s): 668
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?
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.
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
3 4
4 7
4
4 7 8
题解:染色线段树,每次询问输出这段区间的颜色种类;位运算记录颜色;注意lazy传递的是相等,因为直接染色;
代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#define LL long long
using namespace std;
#define ll root<<1
#define rr root<<1|1
#define lson ll, l, mid
#define rson rr, mid + 1, r
int ans;
const int MAXN = 1e6 + ;
int tree[MAXN << ];
int lazy[MAXN << ];
int p[];
void pushup(int root){
tree[root] = tree[ll] | tree[rr];
}
void pushdown(int root){
if(lazy[root]){
lazy[ll] = lazy[root];
lazy[rr] = lazy[root];
tree[ll] = lazy[root];
tree[rr] = lazy[root];
lazy[root] = ;
}
}
void build(int root, int l, int r){
lazy[root] = ;
int mid = (l + r) >> ;
if(l == r){
tree[root] = << ( - );
return;
}
build(lson);
build(rson);
pushup(root);
}
void update(int root, int l, int r, int L, int R, int C){
int mid = (l + r) >> ;
if(l >= L && r <= R){
tree[root] = << (C - );
lazy[root] = << (C - );
return;
}
pushdown(root);
if(mid >= L)update(lson, L, R, C);
if(mid < R)update(rson, L, R, C);
pushup(root);
}
void query(int root, int l, int r, int L, int R){
int mid = (l + r) >> ;
if(l >= L && r <= R){
ans |= tree[root];
return;
}
pushdown(root);
if(mid >= L)query(lson, L, R);
if(mid < R)query(rson, L, R);
}
int main(){
int N, M;
char s[];
while(scanf("%d%d", &N, &M), N|M){
build(, , N);
while(M--){
int l, r, c;
scanf("%s%d%d", s, &l, &r);
if(s[] == 'P'){
scanf("%d", &c);
update(, , N, l, r, c);
}
else{
ans = ;
int tp = , i = ;
query(, , N, l, r);
while(ans){
if(ans & )
p[tp++] = i;
i++;
ans >>= ;
}
for(int i = ; i < tp; i++){
if(i)printf(" ");
printf("%d", p[i]);
}
puts("");
}
}
}
return ;
}
A Corrupt Mayor's Performance Art(线段树区间更新+位运算,颜色段种类)的更多相关文章
- hdu----(5023)A Corrupt Mayor's Performance Art(线段树区间更新以及区间查询)
A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100 ...
- HDU 5023 A Corrupt Mayor's Performance Art 线段树区间更新+状态压缩
Link: http://acm.hdu.edu.cn/showproblem.php?pid=5023 #include <cstdio> #include <cstring&g ...
- 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 way ...
- hdu 5023 A Corrupt Mayor's Performance Art 线段树
A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100 ...
- poj 2777 线段树 区间更新+位运算
题意:有一个长板子,分成多段,有两种操作,第一种是C给从a到b那段染一种颜色c,另一种是P询问a到b有多少种不同的颜色.Sample Input2 2 4 板长 颜色数目 询问数目C 1 1 2P ...
- POJ-2528 Mayor's posters (线段树区间更新+离散化)
题目分析:线段树区间更新+离散化 代码如下: # include<iostream> # include<cstdio> # include<queue> # in ...
- POJ-2528 Mayor's posters(线段树区间更新+离散化)
http://poj.org/problem?id=2528 https://www.luogu.org/problem/UVA10587 Description The citizens of By ...
- POJ 2528 Mayor's posters(线段树/区间更新 离散化)
题目链接: 传送门 Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Description The citizens of By ...
- POJ2528 Mayor's posters(线段树&区间更新+离散化)题解
题意:给一个区间,表示这个区间贴了一张海报,后贴的会覆盖前面的,问最后能看到几张海报. 思路: 之前就不会离散化,先讲一下离散化:这里离散化的原理是:先把每个端点值都放到一个数组中并除重+排序,我们就 ...
随机推荐
- iOS FMDB中的使用
n使用事务 [queue inTransaction:^(FMDatabase *db, BOOL *rollback) { [db executeUpdate:@"INSERT INTO ...
- 【软件技巧】Sublime Text为不同语法定义不同高亮
Sublime Text默认的语法高亮已经非常美丽了,可是对于个别语言还是有些不爽. 默认高亮规则叫Monokai,能够从Preferences->Settings - Default中看到: ...
- C函数的实现(strcpy,atoi,atof,itoa,reverse)
在笔试面试中经常会遇到让你实现C语言中的一些函数比如strcpy,atoi等 1. atoi 把字符串s转换成数字 int Atoi( char *s ) { int num = 0, i = 0; ...
- 【Android】实现动态显示隐藏密码输入框的内容
在设置输入密码框时,有些时候需要按钮控制输入的是“明文”或者“暗文”. 这里提供一种Android实现动态显示隐藏密码输入框的内容的方法: 主要是通过设置EditText的setTransformat ...
- Ubuntu包管理命令 dpkg、apt和aptitude
起初GNU/Linux系统中仅仅有.tar.gz.用户 必须自己编译他们想使用的每个程序.在Debian出现之後,人们觉得有必要在系统 中加入一种机 制用来管理 安装在计算机上的软件包.人们将这套系统 ...
- Linux远程登录
Linux远程登录 远程登录 关闭linux的防火墙 /etc/init.d/iptables stop 启动VNC服务器 Vncserver & 然后记住desktop is localho ...
- MySQL 触发器的定义
-- Insert DELIMITER $$ USE `testdatabase`$$ DROP TRIGGER /*!50032 IF EXISTS */ `Trigger_XXX_INSERT`$ ...
- 提示框的优化之自定义Toast组件之(三)Toast组件优化
开发步骤: 在toast_customer.xml文件中添加一个图片组件对象显示提示图片 <?xml version="1.0" encoding="utf-8&q ...
- MVC 数据列表显示插件大全
Jgrid 官网示例: http://www.trirand.net/demo/aspnet/mvc/jqgrid/ Code Project示例: http://www.codeproject.co ...
- OC——UIlabel text的常规应用
UILabel *downloader = [[UILabel alloc]init]; NSString *downloadCount = [[LibraryArr objectAtIndex:in ...