POJ训练计划2777_Count Color(线段树/成段更新/区间染色)

解题报告

题意：

对线段染色。询问线段区间的颜色种数。

思路：

本来直接在线段树上染色，lz标记颜色。每次查询的话訪问线段树，求出颜色种数。结果超时了，最坏的情况下，染色能够染到叶子节点。

换成存下区间的颜色种数，这样每次查询就不用找到叶子节点了。用按位或来处理颜色种数。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int sum[500000],lz[500000],ans;
void push_down(int rt,int l,int r)
{
    if(lz[rt])
    {
        lz[rt<<1]=lz[rt<<1|1]=lz[rt];
        sum[rt<<1]=lz[rt];
        sum[rt<<1|1]=lz[rt];
        lz[rt]=0;
    }
}
void push_up(int rt,int l,int r)
{
    sum[rt]=sum[rt<<1]|sum[rt<<1|1];
}
void cbtree(int rt,int l,int r)
{
    if(l==r)
    {
        sum[rt]=1;
        return ;
    }
    int mid=(l+r)>>1;
    cbtree(rt<<1,l,mid);
    cbtree(rt<<1|1,mid+1,r);
    push_up(rt,l,r);
}
void update(int rt,int l,int r,int ql,int qr,int v)
{
    if(ql>r||qr<l)return ;
    if(ql<=l&&r<=qr)
    {
        lz[rt]=v;
        sum[rt]=v;
        return ;
    }
    int mid=(l+r)>>1;
    push_down(rt,l,r);
    update(rt<<1,l,mid,ql,qr,v);
    update(rt<<1|1,mid+1,r,ql,qr,v);
    push_up(rt,l,r);
}
int _q(int rt,int l,int r,int ql,int qr)
{
    if(ql>r||qr<l)return 0;
    if(ql<=l&&r<=qr)
    {
        return sum[rt];
    }
    push_down(rt,l,r);
    int mid=(l+r)>>1;
    return _q(rt<<1,l,mid,ql,qr) | _q(rt<<1|1,mid+1,r,ql,qr);

}
int main()
{
    int n,t,q,ql,qr,i,j,a,k;
    char str[10];
    scanf("%d%d%d",&n,&t,&q);
    cbtree(1,1,n);
    while(q--)
    {
        scanf("%s",str);
        if(str[0]=='C')
        {
            scanf("%d%d%d",&ql,&qr,&a);
            if(ql>qr)swap(ql,qr);
            update(1,1,n,ql,qr,1<<(a-1));
        }
        else
        {
            scanf("%d%d",&ql,&qr);
            if(ql>qr)swap(ql,qr);
            ans=_q(1,1,n,ql,qr);
            int cnt=0;
            while(ans)
            {
                if(ans%2)
                    cnt++;
                ans/=2;
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35143		Accepted: 10591

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