POJ - 2777 Count Color（線段樹 + 狀態壓縮）

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

題意：

進行兩種操作：

（1）給區間內塗上某種顏色

（2）求某一區間內不同顏色的數量

思路：

由於顏色種數比較少，可以使用狀態壓縮，用二進制數的每一位代表一種顏色

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
const int inf = 0x3f3f3f3f;

int sum[N << 2], add[N << 2];

void push_up(int rt)
{
    sum[rt] = sum[rt << 1] | sum[rt << 1 | 1];
}

void push_down(int rt)
{
    if(add[rt])
    {
        add[rt << 1] = add[rt << 1 | 1] = add[rt];
        sum[rt << 1] = sum[rt << 1 | 1] = add[rt];
        add[rt] = 0;
    }
}

void build(int l, int r, int rt)
{
    sum[rt] = add[rt] = 0;
    if(l == r)
    {
        sum[rt] = 1;
        return ;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    push_up(rt);
}

void update(int L, int R, int C, int l, int r, int rt)
{
    if(L > r || R < l) return ;
    if(L <= l && R >= r)
    {
        add[rt] = 1 << (C - 1);
        sum[rt] = 1 << (C - 1);
        return ;
    }
    push_down(rt);
    int m = (l + r) >> 1;
    update(L, R, C, l, m, rt << 1);
    update(L, R, C, m + 1, r, rt << 1 | 1);
    push_up(rt);
}

int query(int L, int R, int l, int r, int rt)
{
    if(L > r || R < l) return 0;
    if(L <= l && R >= r) return sum[rt];
    int m = (l + r) >> 1;
    push_down(rt);
    return query(L, R, l, m, rt << 1) | query(L, R, m + 1, r, rt << 1 | 1);
}

int main()
{
    int L, T, O;
    scanf("%d%d%d", &L, &T, &O);
    build(1, L, 1);
    for(int i = 1; i <= O; ++i)
    {
        getchar();
        char op;
        int l, r, k;
        scanf("%c%d%d", &op, &l, &r);
        if(l > r)
            swap(l, r);
        if(op == 'C')
        {
            scanf("%d", &k);
            update(l, r, k, 1, L, 1);
        }
        else
        {
            int ans = query(l, r, 1, L, 1);
            int cnt = __builtin_popcount(ans);///求某個數二進制中1的個數
            printf("%d\n", cnt);
        }
    }
    return 0;
}