Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34950		Accepted: 10542

Description

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. 

There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board: 

1. "C A B C" Color the board from segment A to segment B with color C. 

2. "P A B" Output the number of different colors painted between segment A and segment B (including). 

In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your. 

Input

First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.

Output

Ouput results of the output operation in order, each line contains a number.

Sample Input

2 2 4C 1 1 2P 1 2C 2 2 2P 1 2

Sample Output

21

Source

给出n的长度的木棒，初始的颜色都为1，给出num中颜色，给出m个操作， C l r x 将l到r内的全部颜色更改为x ， P l r 问在l到r内有多少种颜色

用线段树存储下当前的每一段的颜色，更改用数组lazy标记，将颜色转化为二进制数，统计一段颜色时，对每一段能够对它的左右子树取 | 这样就能够统计这一段中的颜色出现的种类。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;#define maxn 110000#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (l+r)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rt#define INF 0x3f3f3f3f#define eqs 1e-6#define LL __int64#define mod 10007#define zero(x) ( fabs(x) < eqs ? 0 : x )#define mem(a,b) (memset(a),b,sizeof(a))int cl[maxn<<2] ;int lazy[maxn<<2] ;void push_up(int_now){ cl[rt] = cl[rt<<1] | cl[rt<<1|1] ;}void push_down(int_now){ if( lazy[rt] ) { lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt] ; cl[rt<<1] = cl[rt<<1|1] = lazy[rt] ; lazy[rt] = 0 ; }}void creat(int_now){ cl[rt] = lazy[rt] = 0 ; if( l != r ) { creat(lson); creat(rson); push_up(now); } else cl[rt] = 1 ;}void update(int ll,int rr,int x,int_now){ if( ll > r || rr < l ) return ; if( ll <= l && r <= rr ) { cl[rt] = lazy[rt] = 1<<(x-1); return ; } push_down(now); update(ll,rr,x,lson); update(ll,rr,x,rson); push_up(now);}int query(int ll,int rr,int_now){ if(ll > r || rr < l) return 0; if(ll <= l && r <= rr) return cl[rt] ; push_down(now); return query(ll,rr,lson) | query(ll,rr,rson);}int ans(int x){ int aa = 0 ; while(x) { if(x & 1) aa++ ; x >>= 1 ; } return aa ;}int main(){ int n , ls , m ; int l , r , x ; char str[10] ; while(~scanf("%d %d %d", &n, &ls, &m)) { creat(root); while(m--) { scanf("%s", str); if(str[0] == 'C') { scanf("%d %d %d", &l, &r, &x); if(l > r) swap(l,r); update(l,r,x,root); } else { scanf("%d %d", &l, &r); if(l > r) swap(l,r); printf("%d\n",ans( query(l,r,root) ) ); } } } return 0;}