P1558 色板游戏

P1558 色板游戏

题目地址
思路：位运算，状态压缩，线段树

1. 分析思路

   • 求某个区间的颜色集合的运算时支持 结合律的 ；
   • 看到 $T\le 30$ 能想到 状态压缩 ；
   • 所以可以设置状态：$\text{X}$的二进制下的第$\text{i}$位为$\text{1}$表示有第$\text{i}$种颜色，否则没有 ；
   • 合并两个状态用到了按位或运算 ；

2. 线段树

   • $\text{Pushup}$操作直接求或运算即可
   • 其他操作照常

3. Code

   #include 
   #include 
   #define Clean(X) memset(X,0,sizeof(X))
   #define LS RT*2
   #define RS RT*2+1
   const int MaxL = 100005 , MaxK = 30;
   unsigned L , K , O ;
   inline unsigned QRead () {
   	unsigned X = 0;
   	char C = getchar() ;
   	while (C > '9' || C < '0') C = getchar() ;
   	while (C >='0' && C <='9') {
   		X = X * 10 + C -'0' ;
   		C = getchar() ;
   	}
   	return X ;
   }
   /*
   头文件和准备工作
   */
   unsigned Cnt(unsigned X) {
   	unsigned Ans = 0 ;
   	while (X) {
   		++Ans ;
   		X -= (X&(-X)) ;
   	}
   	return Ans ;
   }
   /*
   Cnt函数可以统计一个数的二进制位种所含有的1的个数。
   每次进行Lowbit运算，都消除了一个1和它后面的0。
   */
   unsigned Date[MaxL * 4] , Left[MaxL * 4] , Right [MaxL * 4] , Tag[MaxL * 4] ;
   void Plant (unsigned RT , unsigned L , unsigned R) {
   	Left[RT] = L , Right[RT] = R ;
   	Date[RT] = 1 ;
   	Tag[RT] = 0 ;
   	if (L == R) return  ;
   	unsigned Mid = (L + R) >> 1 ;
   	Plant (LS , L , Mid) , Plant (RS , Mid + 1 , R) ;
   }
   inline void Spd (unsigned RT) {
   	Tag[LS] = Tag[RS] = Date[LS] = Date[RS] = Tag[RT];
   	Tag[RT] = 0 ;
   }
   unsigned Ask (unsigned RT , unsigned L , unsigned R) {
   	if (Left[RT] >= L && Right[RT] <= R) return Date[RT] ;
   	unsigned Mid = (Left[RT] + Right[RT] ) >> 1 , Al = 0 , Ar = 0;
   	if (Tag[RT])Spd(RT) ;
   	if (L <= Mid) Al = Ask (LS , L , R) ;
   	if (R >  Mid) Ar = Ask (RS , L , R) ;
   	return Al | Ar ;
   }
   void Add (unsigned RT , unsigned L , unsigned R,unsigned K) {
   	if (Left[RT] >= L && Right[RT] <= R) {
   		Date[RT] = Tag[RT] = K ;
   		return  ;
   	}
   	unsigned Mid = (Left[RT] + Right[RT] ) >> 1 ;
   	if (Tag[RT] )Spd (RT) ;
   	if (L <= Mid) Add (LS , L , R , K) ;
   	if (R >  Mid) Add (RS , L , R , K) ;
   	Date[RT] = Date[LS] | Date[RS] ;
   }
   /*
   线段树的基本函数
   */
   void Swap(unsigned int &X, unsigned int &Y) {
   	int T = X ;
   	X = Y ;
   	Y = T ;
   }
   int main () {
   //	freopen ("P1558.in" , "r" , stdin) ;
   	L = QRead () , K = QRead () , O = QRead () ;
   	Plant (1 , 1 , L) ;
   	for (register unsigned i = 1 ;  i <= O ; ++ i) {
   		char C = getchar() ;
   		while (C != 'P' && C != 'C') C = getchar() ;
   		if (C == 'C') {
   			unsigned L = QRead () , R = QRead () , Cl = QRead () - 1 ;
   			if (L > R) Swap (L , R) ;
   			Add (1 , L , R , 1 << Cl) ;
   		} else {
   			unsigned L = QRead () , R = QRead () ;
   			if (L > R) Swap (L , R) ;
   			printf ("%d\n" , Cnt (Ask (1 , L , R)));
   		}
   	}
   	fclose (stdin) ;
   	fclose (stdout);
   	return 0;
   }
   

Thanks!