POJ2528 Mayor's posters 线段树

POJ 2528 Mayor's posters 线段树

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39344		Accepted: 11424

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
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

Source

Alberta Collegiate Programming Contest 2003.10.18

传送门：POJ 2528 Mayor's posters

吐槽一下这道题 做过一模一样的在计蒜客上 时间也一样 计蒜客的数据用暴力就能过 在POJ上不行 没办法 必须要写个线段树了 

那么这题离散化的时候要注意一个问题 由于他是一个广告版 如果两个点不相邻 那么他中间肯定有一块区域 而在我们的离散化步骤里面 我们必须手动加上一个这个点 代表是这个区域 因为你不加的话 离散化会出问题

怎么加点呢？ 我可以让左端点+1 这样我二分查找的时候是不是找区间单调 所以 满足条件 能用二分查找找原始下标

看query函数 查询的时候返回条件为啥是 col[p]|l==r? l==r好理解 你找道叶子区间了(我们这是区间树)那么叶子区间有颜色 我们就可以cover[col[p]] = 1,代表染上这个色了 如果没颜色呢？没关系 cover[0] = 1，我们求和不加cover[0],

col[p]是啥意思呢？是这区间已经被这个染色了 所以可以直接标记

为啥不down操作？

没意义 你这是覆盖一个颜色 你down下去儿子区间都是这颜色 你就可以直接在爸爸的时候就return了 down也没问题

对时间没有影响

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define lson l , m , o << 1
#define rson m + 1 , r , o << 1 | 1
const int maxN = 10005 ;

int col[maxN << 4] ;
bool cover[maxN] ;
int a[maxN << 2] , b[maxN << 2];
int LL[maxN] , RR[maxN] ;

int Unique ( int *a , int n ) {//离散，返回最后数据的个数
	int cnt = 0 ;
	sort ( a , a + n ) ;
	b[0] = a[0] ;
	for ( int i = 1 ; i < n ; ++ i ) {//unique
		if ( a[i] != b[cnt] ) b[++ cnt] = a[i] ;
	}
	n = cnt + 1 ;
	cnt = 0 ;
	for ( int i = 1 ; i < n ; ++ i) {//添加中间元素
		if ( b[i] != a[cnt] + 1 ) {
			a[cnt + 1] = a[cnt] + 1 ;//确保单调性，以便二分查找
			++ cnt ;
		}
		a[++ cnt] = b[i] ;
	}
	return cnt + 1 ;
}

int Binary_Search ( int *a , int key , int n ) {
	int l = 0 , r = n ;
	while ( l <= r ) {
		int m = ( l + r ) / 2 ;
		if ( key <= a[m] ) r = m-1 ;
		else l = m + 1 ;
	}
	return l ;
}

void down ( int p,int l,int r ) {
	if ( col[p] ) {
		col[p << 1] = col[p << 1 | 1] = col[p] ;
		col[p] = 0 ;
	}
}

void change ( int p,int l,int r,int x,int y,int v ) {
	if ( x<= l && r <= y ) {
		col[p] = v ;
		return ;
	}
	down(p,l,r);
	int mid = ( l + r ) >> 1 ;
	if ( x <= mid ) change ( p*2,l,mid,x,y,v) ;
	if ( y>mid ) change ( p*2+1,mid+1,r,x,y,v) ;
}

void query ( int p,int l , int r  ) {
	if ( col[p] || l == r ) {
		cover[col[p]] = 1 ;
		//printf ( "l = %d , r = %d , col[%d] = %d\n" , l , r , o , col[o] ) ;
		return ;
	}
	//down ( p ) ;很明显不需要了，因为只要不同的颜色，而PushDown下去的必定是同种颜色
	int mid = ( l + r ) >> 1 ;
	query ( p*2,l,mid ) ;
	query ( p*2+1,mid+1,r ) ;
}

void work () {
	int n , l , r , c , cnt , color ;
	scanf ( "%d" , &n ) ;
	memset(col,0,sizeof(col));
	memset(cover,0,sizeof(cover));
	cnt = 0 ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		scanf ( "%d%d" , &LL[i] , &RR[i] ) ;
		a[cnt ++] = LL[i] ;
		a[cnt ++] = RR[i] ;
	}
	cnt = Unique ( a , cnt ) ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		l = Binary_Search ( a , LL[i] , cnt ) ;
		r = Binary_Search ( a , RR[i] , cnt ) ;
		//printf ( "l = %d r = %d\n" , l , r ) ;
		change ( 1,1,cnt,l+1,r+1,i ) ;
	}
	query ( 1,1,cnt) ;
	color = 0 ;
	for ( int i = 1 ; i <= n ; ++ i ) {
		color += cover[i] ;
	}
	printf ( "%d\n" , color ) ;
}
int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		work () ;
	}
	return 0 ;
}