POJ 2777 Count Color

这两天在练习各种线段树，于是就继续更一道线段树的题目QAQ

题面如下：

Count Color

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 49616		Accepted: 14946

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：和之前有个贴海报的题差不多，意思就是给你一个区间和一些颜色，区间初始全为颜色1，C操作代表将某一区间涂成某一种颜色，P操作是询问某一区间的颜色个数。

分析：一道比较裸的线段树，在这儿可以稍微用一下lazy标记。有个思想，如何在线段树中体现两个区间不同颜色的个数？我们这儿采用了二进制存储信息的思想，因为它总共只有30发颜色，所以我们用一个int位的二进制表示就足够了。两个区间的不同颜色个数就是把他们或起来就可以了。

这儿注意比较坑的一点是，查询的区间有可能第一个数是大于第二个数的QAQ

代码如下：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;

const int maxn=1e5+7;
int l,t,o;
int tree[maxn<<2],Hash[maxn],lazy[maxn<<2];

void pushup(int rt){
    tree[rt]=tree[rt<<1]|tree[rt<<1|1];
}

void pushdown(int rt){
    if(lazy[rt]){
        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
        tree[rt<<1]=tree[rt<<1|1]=lazy[rt];
        lazy[rt]=0;
    }
}

void build(int l,int r,int rt){
    tree[rt]=lazy[rt]=0;
    if(l!=r){
        int m=(l+r)>>1;
        build(lson);
        build(rson);
        pushup(rt);
    }
    else tree[rt]=1;
}

void update(int L,int R,int c,int l,int r,int rt){
    if(L>r||R<l) return ;
    if(L<=l&&r<=R){
        tree[rt]=lazy[rt]=1<<(c-1);
        return ;
    }
    pushdown(rt);
    int m=(l+r)>>1;
    update(L,R,c,lson);
    update(L,R,c,rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
    if(L>r||l>R) return 0;
    if(L<=l&&r<=R){
        return tree[rt];
    }
    pushdown(rt);
    int m=(l+r)>>1;
    return query(L,R,lson)|query(L,R,rson);
}

int ans(int c){
    int aa=0;
    while(c){
        if(c&1)
            aa++;
        c>>=1;
    }
    return aa;
}


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>l>>t>>o;
    build(1,l,1);

    char s;

    while(o--){
        cin>>s;
        int x,y,z;
        if(s=='C') {
            cin>>x>>y>>z;
            if(x>y) swap(x,y);
            update(x,y,z,1,l,1);
        }
        else{
            cin>>x>>y;
            if(x>y) swap(x,y);
            cout<<ans(query(x,y,1,l,1))<<endl;
        }
    }
    return 0;
}