poj2777

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 Output21

思路：填色，一开始完全没思路，看了下其他大佬的思路，因为颜色数量只有30可以用二进制来表示颜色，含有1代表有这种颜色，如101，代表有第一三种颜色，共两种。子区间可以通过‘|‘符号（或 的意思）来合并出上一级区间的状态如 101 | 110 = 111 .父区间含有三种颜色。因为查询操作较多，必须用线段树区间更新，主要是延迟标记，而且要用scanf输入.

实现代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
const int M = 1e6+10;
int n,w,e;
int color[M<<2],lazy[M<<2];
void pushup(int rt){
   color[rt] = color[rt*2]|color[rt*2+1];
}

void build(int l,int r,int rt){
    if(l==r){
        color[rt] = 1;
        return ;
    }
    int m = (l+r)/2;
    build(l,m,rt*2);
    build(m+1,r,rt*2+1);
    pushup(rt);
}

void pushdown(int rt){
    if(lazy[rt]){
        lazy[rt*2] = (1<<(lazy[rt]-1));
        lazy[rt*2+1] = (1<<(lazy[rt]-1));
        color[rt*2] = color[rt];
        color[rt*2+1] = color[rt];
        lazy[rt] = 0;
    }
}

void update(int L,int R,int l,int r,int x,int rt){
     if(L<=l&&r<=R){
        lazy[rt] = x;
        color[rt] = (1<<(x-1));
        return ;
     }
     pushdown(rt);
     int m = (l+r)/2;
     if(L<=m)  update(L,R,l,m,x,rt*2);
     if(R>m) update(L,R,m+1,r,x,rt*2+1);
     pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return color[rt];
    }
    int ans1=0,ans2=0,ans;
    int m = (l+r)/2;
    pushdown(rt);
    if(L<=m) ans1+=query(L,R,l,m,rt*2);
    if(R>m)  ans2+=query(L,R,m+1,r,rt*2+1);
    ans = ans1|ans2;
    return ans;
}

void getsum(int x){
    int ans = 0;
    while(x){
        if(x%2==1) ans++;
        x/=2;
    }
    cout<<ans<<endl;
}

int main()
{
    int L,R,x;
    char c;
    while(scanf("%d%d%d",&n,&w,&e)!=EOF){
        memset(lazy,0,sizeof(lazy));
        build(1,n,1);
        while(e--){
        scanf("%c",&c);
        while(c==‘ ‘||c == ‘\n‘) scanf("%c",&c);
        if(c==‘C‘){
            scanf("%d%d%d",&L,&R,&x);
            if(L>R) swap(L,R);
            update(L,R,1,n,x,1);
        }
        else{
            scanf("%d%d",&L,&R);
            if(L>R) swap(L,R);
            int cnt = query(L,R,1,n,1);
            getsum(cnt);
        }
    }
    }
    return 0;
}