题目链接:  poj 2777 Count Color 题目大意:  给出一块长度为n的板,区间范围[1,n],和m种染料 k次操作,C  a  b  c 把区间[a,b]涂为c色,P  a  b 查询区间[a,b]有多少种不同颜色 解题思路:  很明显的线段树的区间插入和区间查询,但是如何统计有多少不同的颜色呢? 如果每个结点数组来存储颜色的种类,空间复杂度很高,而且查询很慢 颜色最多只有30种,可以用位运算中的“按位或|” 颜色也用二进制来处理,和存储: 第一种颜色的二进制表示1 第二种颜色…
题目地址:http://poj.org/problem?id=2777 Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30995   Accepted: 9285 Description Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of pr…
Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 42472   Accepted: 12850 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…
Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38921   Accepted: 11696 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.…
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 33311 Accepted: 10058 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. The…
题目链接:http://poj.org/problem?id=2777 题意是有L个单位长的画板,T种颜色,O个操作.画板初始化为颜色1.操作C讲l到r单位之间的颜色变为c,操作P查询l到r单位之间的颜色有几种. 很明显的线段树成段更新,但是查询却不好弄.经过提醒,发现颜色的种类最多不超过30种,所以我们用二进制的思维解决这个问题,颜色1可以用二进制的1表示,同理,颜色2用二进制的10表示,3用100,....假设有一个区间有颜色2和颜色3,那么区间的值为二进制的110(十进制为6).那我们就把…
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 d…
职务地址:id=2777">POJ 2777 我去.. 延迟标记写错了.标记到了叶子节点上.. . . 这根本就没延迟嘛.. .怪不得一直TLE... 这题就是利用二进制来标记颜色的种类.然后利用或|这个符号来统计每一个区间不同颜色种数. 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.…
题目连接 http://poj.org/problem?id=2777 Count Color 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 centime…
Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 42940   Accepted: 13011 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.…
Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 53639   Accepted: 16153 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.…
传送门:Count Color 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…
题目大意:要求完成以下两个操作:1.将一个区间刷上一种颜色2.询问一段区间上有多少种颜色 思路:这两个操作线段树都可以很迅速的完成,具体做法是:线段树上每个节点存这个线段上的颜色数量,由于颜色数很少,因此可以用二进制存颜色,如果二进制的第N位是1,则该区间存在颜色N,因此一个节点等于其两个子节点颜色的或.最后一个问题就是修改操作是对区间修改,因此需要用lazy_tag的思想(虽然这里不完全是),不然一个一个插节点会TLE #include<cstdio> #include<string.…
Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 39917 Accepted: 12037 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. The…
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…
Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 53312   Accepted: 16050 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.…
Count Color Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status 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 pr…
题目:http://poj.org/problem?id=2777 区间更新,比点更新多一点内容, 详见注释,  参考了一下别人的博客.... 参考博客:http://www.2cto.com/kf/201402/277917.html #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> using namespace std; + ; ]; struct n…
题目链接:http://poj.org/problem?id=2777 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…
题目链接 http://poj.org/problem?id=2777 题意:题意是有L个单位长的画板,T种颜色,O个操作.画板初始化为颜色1.操作C讲l到r单位之间的颜色变为c,操作P查询l到r单位之间的颜色有几种. 这题区间更新很简单但是查询会发现挺麻烦的,主要是不知道怎么判断一个区间到底有几个不同的数. 由于这道题的T比较小才30,1<<30不超过int型于是可以考虑用状态来表示,1表示color1,10表示color2,100表示color3,以此类推. 然后区间更新时注意父节点的变化…
又毁三观了.......虽然题目数据有坑:区间[a,b]可能会有a>b的情况,但是我一开始没有考虑它也能过. 此外莫名其妙的TLE #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <…
发现自己越来越傻逼了.一道傻逼题搞了一晚上一直超时,凭啥子就我不能过??? 然后发现cin没关stdio同步... 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…
题目描写叙述: 长度为L个单位的画板,有T种不同的颜料.现要求按序做O个操作,操作分两种: 1."C A B C",即将A到B之间的区域涂上颜色C 2."P A B".查询[A,B]区域内出现的颜色种类 出现操作2时.请输出答案 PS:初始状态下画板颜色为1 一開始没有想那么好,用int整型位移来取代颜色.还是使用了最传统的bool color[来记录.但是不知道错在了哪里. #include<iostream> #include<cstdio&g…
POJ 2777 Count Color (线段树)   Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29895   Accepted: 8919 Description Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems.…
描述Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.Your task is counting the segments of different colors you can see at last. InputThe first line of each data set contains exactly…
这道题对于我这样的初学者还是有点难度的不过2遍A了还是很开心,下面说说想法-- Count Color Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 40302 Accepted: 12161 Description Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of p…
题目链接: http://poj.org/problem?id=3264 思路分析: 典型的区间统计问题,要求求出某段区间中的极值,可以使用线段树求解. 在线段树结点中存储区间中的最小值与最大值:查询时使用线段树的查询 方法并稍加修改即可进行查询区间中最大与最小值的功能. 代码(线段树解法): #include <limits> #include <cstdio> #include <iostream> using namespace std; ; + ; struct…
Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 34950   Accepted: 10542 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.…
做这道题之前,建议先做POJ 1151  Atlantis,经典的扫描线求矩阵的面积并 参考连接: http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018702.html 线段树辅助——扫描线法计算矩形周长并(轮廓线):http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018687.htmlhttp://blog.csdn.net/ophunter/article/det…
1: #include <stdio.h> 2: #include <string.h> 3: #include <stdlib.h> 4: #include <algorithm> 5: #include <iostream> 6: using namespace std; 7:   8: #define LL(a) a<<1 9: #define RR(a) a<<1|1 10: const int MaxL = 10…