HDU 3911 Black And White 线段树

Black And White

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3328    Accepted Submission(s): 1020

Problem Description

There are a bunch of stones on the beach; Stone color is white or black. Little Sheep has a magic brush, she can change the color of a continuous stone, black to white, white to black. Little Sheep like black very much, so she want to know the longest period of consecutive black stones in a range [i, j].

 

Input

  There are multiple cases, the first line of each case is an integer n(1<= n <= 10^5), followed by n integer 1 or 0(1 indicates black stone and 0 indicates white stone), then is an integer M(1<=M<=10^5) followed by M operations formatted as x i j(x = 0 or 1) , x=1 means change the color of stones in range[i,j], and x=0 means ask the longest period of consecutive black stones in range[i,j]

 

Output

When x=0 output a number means the longest length of black stones in range [i,j].

 

Sample Input

4
1 0 1 0
5
0 1 4
1 2 3
0 1 4
1 3 3
0 4 4

 

Sample Output

1
2
0

 

Source

2011 Multi-University Training Contest 8 - Host by HUST

传送门：HDU 3911 Black And White

题目大意：给你长度为n的一串01序列，m次操作。n,m<=10^5。

操作有两种：
0 L R 查询区间【L，R】内最长的连续1的长度。

1 L R 翻转区间【L，R】内的数（0变1，1变0）。

题目分析：
区间合并水题。
维护每一个区间的最长前缀。最长后缀，最长子序列即可了。

代码例如以下：

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

#define rt l , r , o
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson l , m , ls
#define rson m + 1 , r , rs

const int maxN = 500000 ;

int lmax[2][maxN] , mmax[2][maxN] , rmax[2][maxN] ;
int cha[maxN] ;
int L , R ;

int max ( const int X , const int Y ) {
	return X > Y ?
 X : Y ;
}

int min ( const int X , const int Y ) {
	return X < Y ? X : Y ;
}

void swap ( int &X , int &Y ) {
	int   tmp ;
	tmp =   X ;
	X   =   Y ;
	Y   = tmp ;
}

void Init_PushUp ( int x , int l , int r , int o ) {
	int m = ( l + r ) >> 1 ;
	lmax[x][o] = lmax[x][ls] ;
	if ( lmax[x][ls] == m - l + 1 ) lmax[x][o] += lmax[x][rs] ;
	
	rmax[x][o] = rmax[x][rs] ;
	if ( rmax[x][rs] == r - m ) rmax[x][o] += rmax[x][ls] ;
	
	mmax[x][o] = max ( mmax[x][ls] , mmax[x][rs] ) ;
	mmax[x][o] = max ( mmax[x][o] , rmax[x][ls] + lmax[x][rs] ) ;
}

void PushUp ( int l , int r , int o ) {
	Init_PushUp ( 0 , rt ) ;
	Init_PushUp ( 1 , rt ) ;
}

void change ( int o ) {
	cha[o] ^= 1 ;
	swap ( lmax[0][o] , lmax[1][o] ) ;
	swap ( mmax[0][o] , mmax[1][o] ) ;
	swap ( rmax[0][o] , rmax[1][o] ) ;
}

void PushDown ( int o ) {
	if ( cha[o] ) {
		change ( ls ) ;
		change ( rs ) ;
		cha[o] = 0 ;
	}
}

void Build ( int l , int r , int o ) {
	cha[o] = 0 ;
	if ( l == r ) {
		int x ;
		scanf ( "%d" , &x ) ;
		lmax[x][o] = mmax[x][o] = rmax[x][o] = 1 ;
		lmax[x ^ 1][o] = mmax[x ^ 1][o] = rmax[x ^ 1][o] = 0 ;
		return ;
	}
	int m = ( l + r ) >> 1 ;
	Build ( lson ) ;
	Build ( rson ) ;
	PushUp ( rt ) ;
}

void Update ( int l , int r , int o ) {
	if ( L <= l && r <= R ) {
		change ( o ) ;
		return ;
	}
	PushDown ( o ) ;
	int m = ( l + r ) >> 1 ;
	if ( L <= m ) Update ( lson ) ;
	if ( m <  R ) Update ( rson ) ;
	PushUp ( rt ) ;
}

int Query ( int l , int r , int o ) {
	if ( L <= l && r <= R ) return mmax[1][o] ;
	PushDown ( o ) ;
	int m = ( l + r ) >> 1 ;
	int tmp1 = 0 , tmp2 = 0 , tmp3 = 0 ;
	if ( L <= m ) tmp1 = Query ( lson ) ;
	if ( m <  R ) tmp2 = Query ( rson ) ;
	tmp3 = min ( m - L + 1 , rmax[1][ls] ) + min ( R - m , lmax[1][rs] ) ;
	return max ( max ( tmp1 , tmp2 ) , tmp3 ) ;
}

void work () {
	int n , m , ch ;
	while ( ~scanf ( "%d" , &n ) ) {
		Build ( 1 , n , 1 ) ;
		scanf ( "%d" , &m ) ;
		while ( m -- ) {
			scanf ( "%d%d%d" , &ch , &L , &R ) ;
			if ( 0 == ch ) printf ( "%d\n" , Query ( 1 , n , 1 ) ) ;
			if ( 1 == ch ) Update ( 1 , n , 1 ) ;
		}
	}
}
int main () {
	work () ;
	return 0 ;
}