poj 2777 count color 线段树

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.

 

简而言之，一条一维的线，反复地给某一连续段涂色，后涂的色覆盖先涂的色，要求一边涂一边随时接受问询，回答某一段有多少种颜色。

很明显的线段树题。

几个key point：一、由于颜色数量比较少，利用了2进制来存储数据。cnt【x】的二进制表达中的第i位为1代表这段板上有第i种颜色。

二、lazy数组，当一整块涂色的时候先不急着向下更新子节点，把lazy置成1表示子区间未更新。当需要知道的时候再pushdown一层，还需要的话再继续pushdown。很懒吧233.

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int MAXN=100010;
int cnt[MAXN<<2];
int lazy[MAXN<<2];

void pushup(int rt)
{
    cnt[rt]=cnt[rt<<1]|cnt[rt<<1|1];
 //   printf("cnt=%d %d %d %d
",cnt[rt],cnt[rt<<1],cnt[rt<<1|1],rt<<1|1);
}
void pushdown(int rt)
{
    if(lazy[rt])
    {
        cnt[rt<<1]=cnt[rt];
        cnt[rt<<1|1]=cnt[rt];
        lazy[rt<<1]=1;
        lazy[rt<<1|1]=1;
        lazy[rt]=0;
    }
}
int query(int a,int b,int l,int r,int rt)
{
   // printf("a
");
    //printf("query %d %d %d %d %d
",a,b,l,r,rt);
    if(a<=l&&b>=r) return cnt[rt];
    pushdown(rt);
    int t1=0,t2=0;
    int m=(r+l)>>1;
    if(a<=m) t1=query(a,b,lson);
    if(b>m) t2=query(a,b,rson);
 //   printf("query %d %d %d %d %d t1=%d t2=%d
",a,b,l,r,rt,t1,t2);
    return t1|t2;
}

void update(int a,int b,int c,int l,int r,int rt)
{
   // printf("update %d %d %d %d %d %d
",a,b,c,l,r,rt);
    if(a<=l&&b>=r)
    {
        cnt[rt]=1<<(c-1);
        lazy[rt]=1;
      //  printf("cnt=%d l=%d r=%d c=%d
",cnt[rt],l,r,c);
        return;
    }
    pushdown(rt);
    int m=(l+r)>>1;
    if(a<=m) update(a,b,c,lson);
    if(b>m) update(a,b,c,rson);
    pushup(rt);
}
int main()
{
  //  freopen("in.txt","r",stdin);
  //  freopen("out.txt","w",stdout);
    int len,numtype,oper,a,b,c,tmp,t,ans;
    char type;
    scanf("%d%d%d",&len,&numtype,&oper);
   for(int i=0;i<MAXN<<2;i++)
   {
        cnt[i]=1;
   }
    memset(lazy,0,sizeof(lazy));
    for(int i=1;i<=oper;i++)
    {
        char type[2];
        scanf("%s",type);
       // printf("%c
",type[0]);
        if(type[0]=='C')
        {
            scanf("%d%d%d",&a,&b,&c);
            if(a>b)
            {
                t=a;a=b;b=t;
            }
            update(a,b,c,1,len,1);
        }
        else
        {
            scanf("%d%d",&a,&b);
            if(a>b)
            {
                t=a;a=b;b=t;
            }
            tmp=query(a,b,1,len,1);
            for(ans=0;tmp>=1;tmp>>=1)
            {
          //  printf("tmp=%d
",tmp);
                if(tmp&1) ans++;
            }
            printf("%d
",ans);
        }
    }
    return 0;
}