poj 2777 Count Color (成段更新+區間求和)

poj 2777 Count Color (成段更新+區間求和)

Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 36646 Accepted: 11053

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

題意就是求一段牆壁顏色種類，因為顏色種類最多30，可以用二進制表示顏色的總數。

#include
#include
#include
#include
#include
using namespace std;
#define ll __int64
#define N 100005
struct node
{
    int l,r;
    int s,v,f;   //顏色種類二級制表示，區間顏色、是否需要向下更新
}f[N*3];
void creat(int t,int l,int r)
{
    f[t].l=l;
    f[t].r=r;
    f[t].v=1;
    f[t].f=0;
    f[t].s=1;  //起始顏色為1,可以用二進制表示
    if(l==r)
    {
        return ;
    }
    int tmp=t<<1,mid=(l+r)>>1;
    creat(tmp,l,mid);
    creat(tmp|1,mid+1,r);
}

void update(int t,int l,int r,int v)
{
    int tmp=t<<1,mid=(f[t].l+f[t].r)>>1;
    if(f[t].l==l&&f[t].r==r)
    {
        f[t].v=v;
        f[t].s=(1<mid)
        update(tmp|1,l,r,v);
    else
    {
        update(tmp,l,mid,v);
        update(tmp|1,mid+1,r,v);
    }
    f[t].f=0;          //向上求和
    f[t].s=f[tmp].s|f[tmp|1].s;     
}
int query(int t,int l,int r)
{
    if(f[t].l==l&&f[t].r==r)
    {
        return f[t].s;
    }
    int tmp=t<<1,mid=(f[t].l+f[t].r)>>1;
    if(f[t].f)
    {
        f[tmp].f=f[tmp|1].f=1;
        f[tmp].v=f[tmp|1].v=f[t].v;
        f[tmp].s=f[tmp|1].s=(1<mid)
        return query(tmp|1,l,r);
    else
    {
        return query(tmp,l,mid)|query(tmp|1,mid+1,r);
    }
}
int main()
{
    int n,t,q,l,r,c;
    char ch;
    while(~scanf("%d%d%d",&n,&t,&q))
    {
        creat(1,1,n);
        while(q--)
        {
            getchar();
            scanf("%c ",&ch);
            if(ch=='C')
            {
                scanf("%d%d%d",&l,&r,&c);
                if(l>r)
                    swap(l,r);
                update(1,l,r,c-1);
            }
            else
            {
                scanf("%d%d",&l,&r);
                if(l>r)
                    swap(l,r);
                int tmp=query(1,l,r),ans=0;
                while(tmp)
                {
                    if(tmp&1)
                        ans++;
                    tmp>>=1;
                }
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}