POJ2777(线段树涂色问题)

Count Color

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

思路：因为只有30种颜色，所以可以用2进制数颜色。注意：l可能大于r

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=100005;
typedef long long ll;
struct Node{
    ll color,lazy;
    int l,r;
}a[MAXN*3];
int n,m;
ll res;
void pushUp(int rt)
{
    a[rt].color=a[rt<<1].color | a[(rt<<1)|1].color;
}
void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    a[rt].lazy=0;
    if(l==r)
    {
        a[rt].lazy=1;
        a[rt].color=1;
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build((rt<<1)|1,mid+1,r);
    pushUp(rt);
}
void pushDown(int rt)
{
    if(a[rt].lazy!=0)
    {
        a[rt<<1].lazy=a[rt<<1].color=a[rt].color;
        a[(rt<<1)|1].lazy=a[(rt<<1)|1].color=a[rt].color;
        a[rt].lazy=0;
    }
}
void update(int rt,int l,int r,int val)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        ll e=1;
        e<<=(val-1);
        a[rt].color=e;
        a[rt].lazy=e;
        return ;
    }
    pushDown(rt);
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
    {
        update(rt<<1,l,r,val);
    }
    else if(mid<l)
    {
        update((rt<<1)|1,l,r,val);
    }
    else
    {
        update(rt<<1,l,mid,val);
        update((rt<<1)|1,mid+1,r,val);
    }
    pushUp(rt);
}
void query(int rt,int l,int r)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        res=res|a[rt].color;
        return ;
    }
    pushDown(rt);
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
    {
        query(rt<<1,l,r);
    }
    else if(mid<l)
    {
        query((rt<<1)|1,l,r);
    }
    else
    {
        query(rt<<1,l,mid);
        query((rt<<1)|1,mid+1,r);
    }
}
int main()
{
    while(scanf("%d%*d%d",&n,&m)!=EOF)
    {
        build(1,1,n);
        for(int i=0;i<m;i++)
        {
            scanf("%*c");
            char op;
            int l,r;
            scanf("%c %d %d",&op,&l,&r);
            if(l>r)    swap(l,r);
            if(op==‘C‘)
            {
                int val;
                scanf("%d",&val);
                update(1,l,r,val);
            }
            else
            {
                res=0;
                query(1,l,r);
                int cnt=0;
                while(res>0)
                {
                    if(res&1)    cnt++;
                    res>>=1;
                }
                printf("%d\n",cnt);
            }
        }
    }
    return 0;
}