Poj2777Count Color (line segment tree)

Poj2777Count Color (line segment tree)
Count Color

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:40404		Accepted:12188

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 previusly.

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

POJ Monthly -- 2006.03.26, dodo
A long (L) canvas with T color and O operations. Each operation changes a color range or queries the number of colors contained in a range. Analysis: This question is very similar to hud1698, so the query is different. For details, refer to the code and HDU1698.

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           # Include using namespace std; const double eps = 1e-6; const double pi = acos (-1.0); const int INF = 0x3f3f3f; const int MOD = 1000000007; # define ll long # define CL (a, B) memset (a, B, sizeof (a) # define MAXN 100010 struct node {int l, r, s ;} t [MAXN <2]; int L, O, T; int sum [33]; // you can use an array because there are not many colors; 1 indicates that the range has this color, 0 and vice versa void build (int l, int r, int I) {t [I]. l = l; t [I]. r = r; t [I]. s = 1; if (l = r) return; int mid = (L + r)> 1; build (l, mid, I <1); build (mid + 1, r, I <1 | 1 );} void update (int l, int r, int m, int I) {if (t [I]. s = m) return; if (t [I]. l = l & t [I]. r = r) {t [I]. s = m; return;} if (t [I]. s! =-1) {t [I <1]. s = t [I <1 | 1]. s = t [I]. s; t [I]. s =-1;} int mid = (t [I]. l + t [I]. r)> 1; if (l> mid) update (l, r, m, I <1 | 1); else if (r <= mid) update (l, r, m, I <1); else {update (l, mid, m, I <1); update (mid + 1, r, m, I <1 | 1) ;}} void query (int l, int r, int I) {if (t [I]. s! =-1) // If the solid color marks this color {sum [t [I]. s] = 1; return;} else // otherwise, query the subnode {int mid = (t [I]. l + t [I]. r)> 1; if (l> mid) query (l, r, I <1 | 1); else if (r <= mid) query (l, r, I <1); else {query (l, mid, I <1); query (mid + 1, r, I <1 | 1 );}}} int main () {char ch; int a, B, c; while (scanf ("% d", & L, & T, & O) = 3) {build (1, L, 1); while (O --) {getchar (); scanf ("% c", & ch ); if (ch = 'C') {scanf ("% d", & a, & B, & C); update (a, B, c, 1);} int ans = 0; if (ch = 'P') {CL (sum, 0 ); // scanf ("% d", & a, & B); query (a, B, 1); for (int I = 1; I <= T; I ++) // counts the number of colors that have occurred. if (sum [I]) ans ++; printf ("% d \ n ", ans) ;}}return 0 ;}