(中等) POJ 3225 Help with Intervals , 线段树+集合。
Description
LogLoader, Inc. is a company specialized in providing products for analyzing logs. While Ikki is working on graduation design, he is also engaged in an internship at LogLoader. Among his tasks, one is to write a module for manipulating time intervals, which have confused him a lot. Now he badly needs your help.
In discrete mathematics, you have studied several basic set operations, namely union, intersection, relative complementation and symmetric difference, which naturally apply to the specialization of sets as intervals.. For your quick reference they are summarized in the table below:
Operation Notation Definition
Union A ∪ B {x : x ∈ A or x ∈ B} Intersection A ∩ B {x : x ∈ A and x ∈ B} Relative complementation A − B {x : x ∈ A but x ∉ B} Symmetric difference A ⊕ B (A − B) ∪ (B − A)
Ikki has abstracted the interval operations emerging from his job as a tiny programming language. He wants you to implement an interpreter for him. The language maintains a set S, which starts out empty and is modified as specified by the following commands:
| Command | Semantics |
|---|---|
U T |
S ← S ∪ T |
I T |
S ← S ∩ T |
D T |
S ← S − T |
C T |
S ← T − S |
S T |
S ← S ⊕ T |
#include<iostream>
#include<cstdio>
#include<cstring> #define lson L,M,po*2
#define rson M+1,R,po*2+1 using namespace std; const int N=*; bool COL[*]={};
bool XOR[*]={};
bool vis[]={};
bool have=; void pushDown(int po)
{
if(COL[po])
{
COL[po*]=COL[po*+]=COL[po];
COL[po]=;
XOR[po*]=XOR[po*+]=; // don't forget!!!
} if(XOR[po])
{
XOR[po*]=!XOR[po*];
XOR[po*+]=!XOR[po*+];
XOR[po]=;
}
} void updateC(int ul,int ur,bool type,int L,int R,int po)
{
if(ul>ur)
return; if(ul<=L&&ur>=R)
{
XOR[po]=;
COL[po]=type; return;
} pushDown(po); int M=(L+R)/; if(ul<=M)
updateC(ul,ur,type,lson);
if(ur>M)
updateC(ul,ur,type,rson);
} void updateX(int ul,int ur,int L,int R,int po)
{
if(ul>ur)
return; if(ul<=L&&ur>=R)
{
XOR[po]=!XOR[po]; return;
} pushDown(po); int M=(L+R)/; if(ul<=M)
updateX(ul,ur,lson);
if(ur>M)
updateX(ul,ur,rson);
} void query(int L,int R,int po)
{
if(L==R)
{
vis[L]=COL[po]^XOR[po]; if(vis[L])
have=; return;
} pushDown(po); int M=(L+R)/; query(lson);
query(rson);
} int main()
{
char C;
char t1,t2;
int a,b;
int x,y; while(cin>>C)
{
scanf(" %c%d,",&t1,&a);
scanf("%d%c",&b,&t2); a*=;
b*=; if(t1=='(')
++a;
if(t2==')')
--b; switch(C)
{
case 'U':
updateC(a,b,,,N,);
break; case 'I':
updateC(,a-,,,N,);
updateX(,a-,,N,);
updateC(b+,N,,,N,);
updateX(b+,N,,N,);
break; case 'D':
updateC(a,b,,,N,);
updateX(a,b,,N,);
break; case 'C':
updateC(,a-,,,N,);
updateX(,a-,,N,);
updateC(b+,N,,,N,);
updateX(b+,N,,N,);
updateX(a,b,,N,);
break; case 'S':
updateX(a,b,,N,);
break; }
} query(,N,); if(!have)
{
printf("empty set\n");
return ;
} bool has=;
for(int i=;i<=N+;++i)
{
if(vis[i])
{
if(!has)
{
has=;
if(i%)
printf("(%d,",(i-)/);
else
printf("[%d,",i/);
}
}
else
{
if(has)
{
has=;
if((i-)%)
printf("%d) ",i/);
else
printf("%d] ",(i-)/);
}
}
} return ;
}
(中等) POJ 3225 Help with Intervals , 线段树+集合。的更多相关文章
- poj 3225 Help with Intervals(线段树,区间更新)
Help with Intervals Time Limit: 6000MS Memory Limit: 131072K Total Submissions: 12474 Accepted: ...
- POJ 3225 Help with Intervals --线段树区间操作
题意:给你一些区间操作,让你输出最后得出的区间. 解法:区间操作的经典题,借鉴了网上的倍增算法,每次将区间乘以2,然后根据区间开闭情况做微调,这样可以有效处理开闭区间问题. 线段树维护两个值: cov ...
- (中等) POJ 1436 Horizontally Visible Segments , 线段树+区间更新。
Description There is a number of disjoint vertical line segments in the plane. We say that two segme ...
- POJ 3225 Help with Intervals(线段树)
POJ 3225 Help with Intervals 题目链接 集合数字有的为1,没有为0,那么几种操作相应就是置为0或置为1或者翻转,这个随便推推就能够了,然后开闭区间的处理方式就是把区间扩大成 ...
- POJ.2528 Mayor's posters (线段树 区间更新 区间查询 离散化)
POJ.2528 Mayor's posters (线段树 区间更新 区间查询 离散化) 题意分析 贴海报,新的海报能覆盖在旧的海报上面,最后贴完了,求问能看见几张海报. 最多有10000张海报,海报 ...
- POJ 2528 Mayor's posters(线段树+离散化)
Mayor's posters 转载自:http://blog.csdn.net/winddreams/article/details/38443761 [题目链接]Mayor's posters [ ...
- POJ 2528 Mayor's posters (线段树)
题目链接:http://poj.org/problem?id=2528 题目大意:有一个很上的面板, 往上面贴海报, 问最后最多有多少个海报没有被完全覆盖 解题思路:将贴海报倒着想, 对于每一张海报只 ...
- POJ 2892 Tunnel Warfare(线段树单点更新区间合并)
Tunnel Warfare Time Limit: 1000MS Memory Limit: 131072K Total Submissions: 7876 Accepted: 3259 D ...
- POJ 2777 Count Color(线段树染色,二进制优化)
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 42940 Accepted: 13011 Des ...
随机推荐
- Hibernate 系列教程12-继承-Join策略
Employee public class Employee { private Long id; private String name; HourlyEmployee public class H ...
- linux系统定时重启tomcat
#touch auto-start.sh [root@Linux opt]# echo $LANGen_US.UTF-8 #vim auto-start.sh #!/bin/sh export LAN ...
- android图片加水印,文字
两种方法: 1.直接在图片上写文字 String str = "PICC要写的文字"; ImageView image = (ImageView) this.findViewByI ...
- java内部类继承--构造函数传参
/: innerclasses/InheritInner.java // Inheriting an inner class. class WithInner { class Inner {} } / ...
- img图片inline-block总结
<div style="font-size:0;"> <img data-src="http://image.zhangxinxu.com/image/ ...
- HDU 5722 Jewelry
矩形面积并. 需要转化一下思路:记录每一个位置的数以及位置. 对数字进行从小到大排序,数字一样的按位置从小到大排. 这样,一样的数就在一起了.连续的相同的x个数就可以构成很多解,这些解对应于二维平面上 ...
- VBS脚本随笔
1.定时运行程序与关闭程序的VBS处理方法: do set ws=createobject("wscript.shell") ws.run"你要运行的程序的路径(比如说d ...
- 一个好用简单的布局空间EasyUI
之前项目中都是前端来新写的页面,对于很多后台管理系统来说,新写页面其实比较麻烦. 最近看到一款还是不错的开源页面框架EasyUi http://www.jeasyui.com/index.php 这是 ...
- ntfs mount fail after upgrade win10
http://www.cnblogs.com/wangbo2008/p/3782730.html linux下挂载NTFS分区错误修复 今天在linux下打开win的NTFS硬盘总是提示出错了,而 ...
- filters
http://www.cloudera.com/content/cloudera/en/documentation/core/latest/topics/admin_hbase_filtering.h ...