poj 2777 Count Color 【线段树lazy区间染色 + 查询区间颜色数目 + 状态压缩】

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39950		Accepted: 12050

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

稀里糊涂写了一发，因为没有判断a和b大小，RE两次。
改过之后提交没想到就AC了。。。我感觉有点不科学O__O "… 
路过大牛，若发现代码有bug，请指出，万分感谢！


题意：有一个1——L的区间，现在有T种颜色(T <= 30)和Q次查询。
查询操作
一、C a b c 表示把区间[a, b]染成第c种颜色；
二、P a b 表示查询区间[a, b]有多少种颜色。

思路：因为T最多30，可以考虑用状态压缩记录颜色。利用好状态压缩，下面就是线段树区间更新 + 区间查询了。



AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 100000+10
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
struct Tree
{
    int l, r;
    int color;//记录区间颜色 用二进制表示
    int lazy;//表示该区间是否被同一种颜色覆盖
};
Tree tree[MAXN<<2];
void build(int o, int l, int r)
{
    tree[o].l = l, tree[o].r = r;
    tree[o].color = 1; tree[o].lazy = 1;
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}
void PushDown(int o)
{
    if(tree[o].lazy)
    {
        tree[ll].lazy = tree[rr].lazy = tree[o].lazy;
        tree[ll].color = tree[rr].color = tree[o].color;
        tree[o].lazy = 0;
    }
}
void PushUp(int o)
{
    tree[o].color = tree[ll].color | tree[rr].color;
}
void update(int o, int L, int R, int c)
{
    if(L <= tree[o].l && R >= tree[o].r)
    {
        tree[o].color = (1<<(c-1));
        tree[o].lazy = 1;
        return ;
    }
    PushDown(o);
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(R <= mid)
        update(ll, L, R, c);
    else if(L > mid)
        update(rr, L, R, c);
    else
    {
        update(ll, L, mid, c);
        update(rr, mid+1, R, c);
    }
    PushUp(o);
}
int query(int o, int L, int R)
{
    if(L <= tree[o].l && R >= tree[o].r)
        return tree[o].color;
    PushDown(o);
    int mid = (tree[o].l + tree[o].r) >> 1;
    int sum = 0;
    if(R <= mid)
        return sum | query(ll, L, R);
    else if(L > mid)
        return sum | query(rr, L, R);
    else
        return sum | query(ll, L, mid) | query(rr, mid+1, R);
}
int main()
{
    int L, T, Q;
    while(scanf("%d%d%d", &L, &T, &Q) != EOF)
    {
        build(1, 1, L);
        while(Q--)
        {
            int a, b, c, t;
            char op[10];
            scanf("%s", op);
            if(op[0] == 'C')
            {
                scanf("%d%d%d", &a, &b, &c);
                if(a > b)//注意判断a和b大小
                {
                    t = a;
                    a = b;
                    b = t;
                }
                update(1, a, b, c);
            }
            else
            {
                scanf("%d%d", &a, &b);
                if(a > b)
                {
                    t = a;
                    a = b;
                    b = t;
                }
                int sum = query(1, a, b);
                int ans = 0;
                while(sum)//统计颜色数目
                {
                    if(sum & 1)
                        ans++;
                    sum >>= 1;
                }
                printf("%d\n", ans);
            }
        }
    }
    return 0;
}