zoukankan      html  css  js  c++  java
  • POJ 2777 Count Color

    C - Count Color
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    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

    有两种方法:一种是每次查询时都要统计相应区间延迟标记上颜色的种类,能够用set或简单哈希来实现。

    一种是用二进制表示相应的区间涂了第几种颜色,这样每一个区间除了延迟标记外,能够再开一个数组统计当前涂了哪几种颜色。

    这样就和一般的线段树一样了。

    最后再统计一下1的数目。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #define maxn 100005
    #define inf 0x3f3f3f3f
    typedef long long LL;
    int sum[maxn<<2],col[maxn<<2],ll[maxn<<2],rr[maxn<<2];
    inline void pushup(int i){
        sum[i]=sum[i<<1]|sum[i<<1|1];
    }
    inline void pushdown(int i){
        if(col[i]){
            col[i<<1]=col[i<<1|1]=col[i];
            sum[i<<1]=col[i];
            sum[i<<1|1]=col[i];
            col[i]=0;
        }
    }
    void build(int l,int r,int i){
        ll[i]=l;
        rr[i]=r;
        col[i]=1;
        if(l==r)return;
        int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
        build(l,m,ls);
        build(m+1,r,rs);
        pushup(i);
    }
    void update(int l,int r,int v,int i){
        if(l<=ll[i]&&rr[i]<=r){
            col[i]=1<<(v-1);
            sum[i]=col[i];
            return ;
        }
        pushdown(i);
        int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
        if(l<=m)update(l,r,v,ls);
        if(m<r)update(l,r,v,rs);
        pushup(i);
    }
    int query(int l,int r,int i){
        if(l<=ll[i]&&rr[i]<=r){
            return sum[i];
        }
        pushdown(i);
        int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
        int ans=0;
        if(l<=m)ans=ans|query(l,r,ls);
        if(m<r)ans=ans|query(l,r,rs);
        return ans;
    }
    int main()
    {
        int l,t,o,a,b,c;
        char q[2];
        //freopen("in.txt","r",stdin);
        while(~scanf("%d%d%d",&l,&t,&o)){
            build(1,l,1);
            for(int i=0;i<o;i++){
                scanf("%s%d%d",q,&a,&b);
                if(a>b)swap(a,b);
                if(q[0]=='C'){
                    scanf("%d",&c);
                    update(a,b,c,1);
                }
                else {
                    int res=query(a,b,1);
                    int ans=0;
                    while(res){
                        if(res&1)ans++;
                        res=res>>1;
                    }
                    printf("%d
    ",ans);
                }
            }
        }
    }

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #define maxn 100005
    #define inf 0x3f3f3f3f
    typedef long long LL;
    int ll[maxn<<2],rr[maxn<<2],col[maxn<<2],vis[32];
    int ans;
    inline void pushdown(int i,int m){
        if(col[i]){
            col[i<<1]=col[i<<1|1]=col[i];
            col[i]=0;
        }
    }
    void build(int l,int r,int i){
       ll[i]=l;
       rr[i]=r;
       col[i]=1;
       if(l==r){
         return ;
       }
       int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
       build(l,m,ls);
       build(m+1,r,rs);
    }
    void update(int l,int r,int v,int i){
       if(l<=ll[i]&&rr[i]<=r){
        col[i]=v;
        return ;
       }
       pushdown(i,rr[i]-ll[i]+1);
       int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
       if(l<=m)update(l,r,v,ls);
       if(m<r) update(l,r,v,rs);
    }
    void query(int l,int r,int i){
       if(col[i]){
          if(!vis[col[i]]){
            ans++;
            vis[col[i]]=1;
          }
          return ;
       }
       pushdown(i,rr[i]-ll[i]+1);
       int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
       if(l<=m)query(l,r,ls);
       if(m<r)query(l,r,rs);
    }
    int main()
    {
        int l,t,o,a,b,c;
        char q[2];
        //freopen("in.txt","r",stdin);
        while(~scanf("%d%d%d",&l,&t,&o)){
            build(1,l,1);
            for(int i=0;i<o;i++){
                scanf("%s%d%d",q,&a,&b);
                if(a>b)swap(a,b);
                if(q[0]=='C'){
                    scanf("%d",&c);
                    update(a,b,c,1);
                }
                else {
                    ans=0;
                    memset(vis,0,sizeof vis);
                    query(a,b,1);
                    printf("%d
    ",ans);
                }
            }
        }
    }



  • 相关阅读:
    C++笔记
    mongodb文件损坏的恢复--无可恢复数据
    wiredtiger--初学数据恢复
    mongodb异常恢复
    守护进程,互斥锁, IPC ,Queue队列,生产消费着模型
    进程,多进程,进程与程序的区别,程序运行的三种状态,multiprocessing模块中的Process功能,和join函数,和其他属性,僵尸与孤儿进程
    socket之UDP协议,并发编程介绍,操作系统发展史
    半连接数,粘包问题,自定义报头
    socket模块
    网络编程介绍,C/S 架构,网络通讯协议,osi七层
  • 原文地址:https://www.cnblogs.com/claireyuancy/p/6760802.html
Copyright © 2011-2022 走看看