Mayor's posters

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules:

  • Every candidate can place exactly one poster on the wall.
  • All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
  • The wall is divided into segments and the width of each segment is one byte.
  • Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l i, l i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed.

The picture below illustrates the case of the sample input. 

Sample Input

1
/*
题意:给横坐标轴染色,每次给一个区间内的染色,每染一次会把上一次的覆盖,经过n次染色只会,问你从最上面看,横坐标上总共有几种颜色 初步思路:区间染色问题,就是一个区间set问题,然后最后的查询的时候,记录一下就行了 #超内存:需要离散化
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n,l[],r[];
int t;
bool color[];//用于记录颜色
int res=;
int X[];
int k;
int kk;
/****************************线段树基础模板*********************************/ const int maxn=+;
#define lson i*2, l, m
#define rson i*2+1, m+1, r struct Segtree{ int setv[maxn<<]; void PushDown(int i)
{
if(setv[i]>){
setv[i*]=setv[i*+]=setv[i];
setv[i]=-;//向下更新完了就没有继续用的意义了
}
} void build(int i,int l,int r)
{
// cout<<l<<" "<<r<<" "<<i<<endl;
setv[i]=;
if(l==r)
return ;
int m=(l+r)>>;
build(lson);
build(rson);
}
void query(int ql,int qr,int i,int l,int r)
{
if(ql<=l&&r<=qr){
if(setv[i]>){
color[setv[i]]=true;
return ;
}
}
if(l==r) return ;
PushDown(i);
int m=(l+r)>>;
if(ql<=m) query(ql,qr,lson);
if(m<qr) query(ql,qr,rson);
} void update(int ql,int qr,int val,int i,int l,int r)
{
// cout<<l<<" "<<r<<endl;
if(ql<=l&&r<=qr)
{
setv[i]=val;
return ;
}
PushDown(i);
int m=(l+r)>>;
if(ql<=m) update(ql,qr,val,lson);
if(m<qr) update(ql,qr,val,rson);
}
};
Segtree segtree;
/****************************线段树基础模板*********************************/
void init(){
memset(color,false,sizeof color);
res=;
k=;
kk=;
}
int findx(int key){
int l=,r=k,mid;
while(l<=r){
mid=(l+r)>>;
if(X[mid]==key)
return mid;
else if(X[mid]>key)
r=mid-;
else l=mid+;
}
return ;
}
int main(){
// freopen("in.txt","r",stdin);
scanf("%d",&t);
while(t--){
init();
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d%d",&l[i],&r[i]);
X[kk++]=l[i];
X[kk++]=r[i];
}
sort(X+,X+kk+);
for(int i=;i<kk;i++){//离散化
if(X[i]!=X[i-])
X[++k]=X[i];
}
segtree.build(,,k);
for(int i=;i<=n;i++){
int L=findx(l[i]);
int R=findx(r[i]);
segtree.update(L,R,i,,,k);
}
segtree.query(,k,,,k);
for(int i=;i<=n;i++){
if(color[i]) res++;
}
printf("%d\n",res);
}
return ;
}
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4
 

Mayor's posters的更多相关文章

  1. POJ 2528 Mayor's posters

    Mayor's posters Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Sub ...

  2. 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 [ ...

  3. poj 2528 Mayor's posters(线段树+离散化)

    /* poj 2528 Mayor's posters 线段树 + 离散化 离散化的理解: 给你一系列的正整数, 例如 1, 4 , 100, 1000000000, 如果利用线段树求解的话,很明显 ...

  4. 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 ...

  5. Mayor's posters(线段树+离散化POJ2528)

    Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 51175 Accepted: 14820 Des ...

  6. poj2528 Mayor's posters(线段树之成段更新)

    Mayor's posters Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 37346Accepted: 10864 Descr ...

  7. POJ 2528 Mayor's posters(线段树区间染色+离散化或倒序更新)

    Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 59239   Accepted: 17157 ...

  8. poj-----(2528)Mayor's posters(线段树区间更新及区间统计+离散化)

    Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 43507   Accepted: 12693 ...

  9. Poj 2528 Mayor's posters 分类: Brush Mode 2014-07-23 09:12 84人阅读 评论(0) 收藏

    Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40570   Accepted: 11798 ...

  10. POJ 2528 Mayor’s posters

    Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37982   Accepted: 11030 ...

随机推荐

  1. Ubuntu16.04安装piwik3.0.1

    1.安装PHP环境 sudo apt-get install php7.0-fpm   2.下载piwik3.0.1 https://piwik.org/download/ 下载后解压到/var/ww ...

  2. JavaScript一些常用方法一

    整理以前的笔记,在学习JavaScript时候,经常会用到一些方法,但是有时忘掉了具体用法,因此记下.方便以后查阅. 这篇博文先说明这些方法的用途: splice().push().pop() .sh ...

  3. [转]IOS开发中的CGFloat、CGPoint、CGSize和CGRect

    http://developer.apple.com/library/ios/#documentation/GraphicsImaging/Reference/CGGeometry/Reference ...

  4. 如何用kaldi做孤立词识别-初版

    ---------------------------------------------------------------------------------------------------- ...

  5. java 反射 类装载器

    前言: java语言允许通过程序化的方式间接对Class进行操作,Class文件由类装载器装载后,在jvm中将形成一份描述Class结构的元信息对象,通过该元信息对象可以获知Class的结构信息,如构 ...

  6. Hive 存储类型 StoreType

    file_format: : SEQUENCEFILE | TEXTFILE -- (Default, depending on hive.default.fileformat configurati ...

  7. Hbase对时,时差范围的确定

    Hbase对时具有严格的要求,集群内部所有机器之间的时差默认不能超过30秒,也就是说,一旦某个regionserver节点上的时间与master节点上的时间差值超过30秒,就会导致相应的regions ...

  8. jQuery点击按钮实现div的隐藏和显示切换效果

    <script type="text/javascript"> $(function(){ $('#click_event').click(function(){  i ...

  9. Django创建通用视图函数

    想在我们有两个视图: def thinkingview(request): user = request.user if request.method == 'GET': return render( ...

  10. 使用olami sdk实现一个语音查询股票的iOS程序

    前言 在目前的软件应用中,输入方式还是以文字输入方式为主,但是语音输入的方式目前应用的越来越广泛.在这里介绍一个使用 Olami SDK 编写的一个使用语音输入查询股票的APP Olami SDK的介 ...