Count Color

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

Sample Output

2
1


题目大意：操作是给给区间涂色（颜色种类小于30），询问指定区间颜色有多少种。

用状态压缩的方式记录颜色，然后就是比较标准的线段树了……

带lazy标记的线段树，其实也不难嘛……


#include <iostream>
#include <string>
#include <stdio.h>
#include <cstring>
using namespace std;
struct ty{
	int col;
	bool lazy; 
}tree[600010];
int  t, o, l, sum;

void build (int p, int l, int r)
{
	tree[p].col = 1;
	tree[p].lazy = 1;
	if (l == r) return;
	int mid = (l + r) / 2;
	build(p * 2, l, mid);
	build(p * 2 + 1, mid + 1, r);
}
void push(int p)  //标记向下传
{
	tree[p * 2].col = tree[p].col;
	tree[p * 2 + 1].col = tree[p].col;
	tree[p * 2].lazy = tree[p].lazy;
	tree[p * 2 + 1].lazy = tree[p].lazy;
	tree[p].lazy = 0;
}
void change(int p, int l, int r, int x, int y, int color)
{
	if (x <= l && r <= y)                    //整个区间都要变——改动lazy标记即可
	{
		tree[p].col = color;
		tree[p].lazy = 1;
		return; 
	}
	if (tree[p].col == color) return ;
	if (tree[p].lazy) push(p);                   //传标记
	int mid = (l + r) / 2;
	if (y <= mid) change(p * 2, l, mid, x, y, color);
	else if (x > mid) change(p*2+1, mid+1, r, x, y, color);
	else {
		change(p * 2, l, mid, x, mid, color);
		change(p*2+1, mid+1, r, mid + 1, y, color);
	}
	tree[p].col = tree[p*2].col | tree[p*2+1].col;
}
	
void find(int p, int l, int r, int x,int y)
{
	if (x <= l && r <= y) { sum |= tree[p].col; return;}
	if (tree[p].lazy) {sum |= tree[p].col; return;}
	int mid = (l + r) / 2;
	if (y <= mid) find(p*2,l,mid,x,y);	
	else if (x > mid) find(p*2+1,mid+1,r,x,y);
	else {
		find(p*2,l,mid,x,mid);	
		find(p*2+1,mid+1,r,mid+1,y);	
	}
}
int calc()
{
    int ans = 0;
    while(sum){
        if(sum & 1)
            ans++;
        sum >>= 1;
    }
    return ans;
}

int main()
{
	while (scanf("%d%d%d", &l, &t, &o) != EOF)
	{

		build(1, 1, l);
		for (int i = 1; i <= o; i++)
		{
			char c;
			int x, y, z;
			scanf(" %c", &c);
			if (c == 'P') 
			{
				scanf("%d%d", &x, &y);
				sum = 0;
				if (x > y) {int t = x; x = t; t=y;}
				find(1, 1, l, x, y);
				printf("%d\n",calc());
				
			}
			else{
				scanf("%d%d%d", &x, &y, &z);
				if (x > y) {int t = x; x = t; t=y;}
				change(1, 1, l, x, y, 1 << (z - 1));
			}
		}
	}
	return 0;
}