poj 2777 线段树

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40116		Accepted: 12103

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
题意：一段长度材料需要涂色，找一定区间内的颜色种类；
第一行输入三个非负整数，Ｌ　Ｔ　Ｏ　分别代表长度，颜色种类和下面再输入几行；
下面Ｏ行：＇Ｃ＇　Ａ　Ｂ　Ｃ：表示在Ａ　Ｂ　区间图一种颜色　Ｃ；
　　　　　＇Ｐ＇　Ａ　Ｂ：让你输出Ａ　Ｂ区间内有多少种颜色；
思路：1.延迟覆盖的操作。2.位操作，用 | 来合并颜色种类。3.updata操作时递归回来，两个子节点的信息对父节点的更新。
经典的线段树。

#include <iostream>
#include <stdio.h>
using namespace std;
struct T
{
    int l,r,add,color;
}tree[100010*4];
int color1;
void pushup(int k)
{
    tree[k].color=tree[k<<1].color|tree[k<<1|1].color;
}
void creat(int l,int r,int k)
{
    tree[k].l=l;
    tree[k].r=r;
    tree[k].add=0;
    tree[k].color=1;
    if(l==r)
        return ;
    int mid=(r+l)>>1;
    creat(l,mid,k<<1);
    creat(mid+1,r,k<<1|1);
}
void pushdown(int k)
{
    int x=k*2;
    tree[x].add=1;
    tree[x+1].add=1;
    tree[x].color=tree[k].color;
    tree[x+1].color=tree[k].color;
    tree[k].add=0;
}
void Search(int l,int r,int color,int k)
{
    if(tree[k].l>r||tree[k].r<l)
        return ;
    if(l<=tree[k].l&&r>=tree[k].r)
    {
        tree[k].color=color;
        tree[k].add=1;
        return ;
    }
    if(tree[k].add)
        pushdown(k);
    int mid=(tree[k].r+tree[k].l)>>1;
    if(r<=mid)
        Search(l,r,color,k<<1);
    else if(l>mid)
            Search(l,r,color,k<<1|1);
    else
    {
        Search(l,mid,color,k<<1);
        Search(mid+1,r,color,k<<1|1);
    }
    pushup(k);
}
void p(int l,int r,int k)
{
     if(tree[k].l>r||tree[k].r<l)
        return ;
     if(l<=tree[k].l&&r>=tree[k].r)
     {
         color1|=tree[k].color;
         return;
     }
     if(tree[k].add)
        pushdown(k);
    int mid=(tree[k].r+tree[k].l)>>1;
    if(r<=mid)
        p(l,r,k<<1);
    else if(l>mid)
            p(l,r,k<<1|1);
    else
    {
        p(l,mid,k<<1);
        p(mid+1,r,k<<1|1);
    }
}
void swap(int &a,int &b)
{
    int t=a;a=b;b=t;
}
int main()
{
    int n,t,o,color,r,l;
    scanf("%d%d%d",&n,&t,&o);
    char str[5];
    creat(1,n,1);
    while(o--)
    {
        scanf("%s",str);
        if(str[0]=='C')
        {
            scanf("%d%d%d",&l,&r,&color);
            if(l>r)
                swap(l,r);
            Search(l,r,1<<(color-1),1);
        }
        else
        {
            scanf("%d%d",&l,&r);
            if(l>r) swap(l,r);
            color1=0;
            p(l,r,1);
            int ans=0;
            while(color1)
            {
                if(color1&1)
                    ans++;
                color1>>=1;
            }
            printf("%d
",ans);
        }
    }
    return 0;
}