POJ：2777-Count Color（线段树+状压）

Count Color

Time Limit: 1000MS Memory Limit: 65536K

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:

• “C A B C” Color the board from segment A to segment B with color C.
• “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

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解题心得：

1. 一看就是线段树，但是还加了状压，每一段区间用一个二进制来表示其涂抹情况，假如二进制的第一位为1代表第一种颜色涂抹了，然后父节点等于两个子节点的状态或起来，询问的区间用0去线段树中或，然后数1的个数就行了。区间更新的时候lazy标记一下，但是要注意的是颜色是直接覆盖。
2. 读题眼贱了一下，没看到给的left端和right端可能是交换的，然后一直Runtime Error，怀疑人生啊。

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#include<stdio.h>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 1e6+100;
struct Bitree
{
    int l,r;
    int co;
} bitree[maxn<<2];
int lazy[maxn<<2];
int n,a,b,c,o,t,ans;

void updata(int rt)//向上更新
{
   bitree[rt].co = (bitree[rt<<1].co)|(bitree[rt<<1|1].co);//字节点或起来
}

void build_tree(int rt,int l,int r)//先初始化一棵树
{
    bitree[rt].l = l;
    bitree[rt].r = r;
    bitree[rt].co |= 2;
    if(r == l)
        return ;
    int mid = (l + r)>>1;
    build_tree(rt<<1,l,mid);
    build_tree(rt<<1|1,mid+1,r);
    updata(rt);
}

void pushdown(int rt)//向下更新，注意lazy标记的转移方式就可以了
{
    if(bitree[rt].l == bitree[rt].r || !lazy[rt])
        return ;
    bitree[rt<<1].co =  bitree[rt<<1|1].co = 0;
    bitree[rt<<1].co |= (1<<lazy[rt]);
    bitree[rt<<1|1].co |= (1<<lazy[rt]);
    lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];
    lazy[rt] = 0;
}

void make_lazy(int rt,int l,int r,int L,int R)//区间更新，lazy标记
{
    pushdown(rt);
    if(l == L && r == R)
    {
        lazy[rt] = c;
        bitree[rt].co = 0;
        bitree[rt].co |= (1<<c);
        return ;
    }
    int mid = (L + R) >> 1;
    if(mid >= r)
        make_lazy(rt<<1,l,r,L,mid);
    else if(mid < l)
        make_lazy(rt<<1|1,l,r,mid+1,R);
    else
    {
        make_lazy(rt<<1,l,mid,L,mid);
        make_lazy(rt<<1|1,mid+1,r,mid+1,R);
    }
    updata(rt);
}

void query(int rt,int l,int r,int L,int R)
{
    pushdown(rt);
    if(L == l && R == r)
    {
        ans |= bitree[rt].co;//查询的时候用ans去将线段树中的状态或出来
        return ;
    }
    int mid = (L + R) >> 1;
    if(mid >= r)
        query(rt<<1,l,r,L,mid);
    else if(mid < l)
        query(rt<<1|1,l,r,mid+1,R);
    else
    {
        query(rt<<1,l,mid,L,mid);
        query(rt<<1|1,mid+1,r,mid+1,R);
    }
    updata(rt);
}

int SUM(int x)
{
    int sum = 0;
    while(x)
    {
        if(x&1)
            sum++;
        x >>= 1;
    }
    return sum;
}

int main()
{
    while(cin>>n>>t>>o)
    {
        memset(lazy,0,sizeof(lazy));
        memset(bitree,0,sizeof(bitree));
        build_tree(1,1,n);
        while(o--)
        {
            char s[10];
            scanf("%s",s);
            if(s[0] == 'C')
            {
                scanf("%d%d%d",&a,&b,&c);
                if(a>b)//注意交换啊，坑死了
                    swap(a,b);
                make_lazy(1,a,b,1,n);
            }
            else if(s[0] == 'P')
            {
                ans = 0;
                scanf("%d%d",&a,&b);
                if(a>b)
                    swap(a,b);
                query(1,a,b,1,n);
                ans = SUM(ans);//数最终有多少个1的个数
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}