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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 , rconst 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 ;
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