【HDU】5023 A Corrupt Mayor's Performance Art 线段树

传送门:【HDU】5023 A Corrupt Mayor's Performance Art


题目分析:水水的线段树,首先颜色只有30种,所以状压就好了,然后每次查询就把区间内所有的颜色“或”出来,用位运算判断一下有哪些颜色就好了。。


代码如下:


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

typedef long long LL ;

#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , n
#define rt o , l , r

const int MAXN = 1000005 ;

int cover[MAXN << 2] ;
int color[MAXN << 2] ;
int n , q ;

void pushup ( int o ) {
	color[o] = color[ls] | color[rs] ;
}

void pushdown ( int o ) {
	if ( cover[o] ) {
		cover[ls] = cover[o] ;
		cover[rs] = cover[o] ;
		color[ls] = color[rs] = cover[o] ;
		cover[o] = 0 ;
	}
}

void update ( int L , int R , int v , int o , int l , int r ) {
	if ( L <= l && r <= R ) {
		cover[o] = 1 << v ;
		color[o] = 1 << v ;
		return ;
	}
	int m = mid ;
	pushdown ( o ) ;
	if ( L <= m ) update ( L , R , v , lson ) ;
	if ( m <  R ) update ( L , R , v , rson ) ;
	pushup ( o ) ;
}

int query ( int L , int R , int o , int l , int r ) {
	if ( L <= l && r <= R ) return color[o] ;
	int m = mid ;
	pushdown ( o ) ;
	if ( R <= m ) return query ( L , R , lson ) ;
	if ( m <  L ) return query ( L , R , rson ) ;
	return query ( L , R , lson ) | query ( L , R , rson ) ;
}

void solve () {
	char s[5] ;
	int x , y , c ;
	clr ( color , 0 ) ;
	clr ( cover , 0 ) ;
	update ( 1 , n , 1 , root ) ;
	while ( q -- ) {
		scanf ( "%s%d%d" , s , &x , &y ) ;
		if ( s[0] == 'Q' ) {
			int S = query ( x , y , root ) ;
			int flag = 0 ;
			rep ( i , 0 , 30 ) {
				if ( S & ( 1 << i ) ) {
					if ( flag ) printf ( " " ) ;
					flag = 1 ;
					printf ( "%d" , i + 1 ) ;
				}
			}
			printf ( "\n" ) ;
		} else {
			scanf ( "%d" , &c ) ;
			update ( x , y , c - 1 , root ) ;
		}
	}
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &q ) && ( n || q ) ) solve () ;
	return 0 ;
}


你可能感兴趣的:(HDU)