POJ 2777 Count Color （線段樹區間更新加查詢）

POJ 2777 Count Color （線段樹區間更新加查詢）

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

看了網上不少的題解，感覺沒有那麼麻煩。。就是區間更新，每次統計數量的時候哈希一下就好，詳見代碼。

#include 
#include 
#include 
#include 
#include 
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef long long LL;
const int MAX = 0x3f3f3f3f;
const int maxn = 100010;
int n, a, b, c, t, q, ans;
char s[3];
int cnt[maxn<<2], vis[35];
void down(int o) {
    if(cnt[o]) {
        cnt[o<<1] = cnt[o<<1|1] = cnt[o];
        cnt[o] = 0;
    }
}
void build(int o, int l, int r) {
    cnt[o] = 1;
    if(l == r) return;
    int m = (l+r) >> 1;
    build(lson);
    build(rson);
}
void update(int o, int l, int r) {
    if(a <= l && r <= b) {
        cnt[o] = c;
        return;
    }
    down(o);
    int m = (l+r) >> 1;
    if(a <= m) update(lson);
    if(m < b ) update(rson);
}
void query(int o, int l, int r) {
    if(cnt[o]) {
        if(vis[ cnt[o] ] == 0) {
            ans ++;
            vis[ cnt[o] ] = 1;
        }
        return ;
    }
    int m = (l+r) >> 1;
    if(a <= m) query(lson);
    if(m < b ) query(rson);
}
int main()
{
    scanf("%d%d%d", &n, &t, &q);
    build(1, 1, n);
    while(q--) {
        scanf("%s%d%d", s, &a, &b);
        if(a > b) swap(a, b);
        if(s[0] == 'C') {
            scanf("%d", &c);
            update(1, 1, n);
        } else {
            memset(vis, 0, sizeof(vis));
            ans = 0;
            query(1, 1, n);
            printf("%d\n", ans);
        }
    }
    return 0;
}