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…

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…

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…

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…

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.…

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.…

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: 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…

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.…

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.…

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: 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…

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.…

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.…

这个是一个经典线段树染色问题,不过题目给的是左右左右坐标,即[0,3]包含0-1这一段 1-2这一段 2-3这一段,和传统的染色不太一样,不过其实也不用太着急. 我们把左边的坐标+1,即可,那么[0,3]其实变成了[1,3]而线段树是按照点询问的,也就是每个点代表的颜色,我们就有了1,2,3,这个三个,并且避免了线段树编号不能到0的情况,然后代码就十分简单了,无需laze标记,因为每个节点的颜色就可以当成laze标记,然后不断往下pushdown既可以,当时还有一个问题就是线段树访问连续两个节点…

D. Slalom time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Little girl Masha likes winter sports, today she's planning to take part in slalom skiing. The track is represented as a grid comp…

[BZOJ3672][NOI2014]购票(线段树,斜率优化,动态规划) 题解 首先考虑\(dp\)的方程,设\(f[i]\)表示\(i\)的最优值 很明显的转移\(f[i]=min(f[j]+(dep[i]-dep[j])·p[i])+q[i]\) 其中满足\(dep[i]-dep[j]\le L[i]\) 然后就可以写出一个\(O(n^2)\)的做法啦 #include<iostream> #include<cstdio> #include<cstdlib> #in…

我会告诉你我看了很久很久才把题目看懂吗???怀疑智商了 原来他给的l,r还有k个数字都是下标... 比如给了一个样例 l, r, k, x1,x2,x3...xk,代表的是一个数组num[l]~num[r],其中有k个数num[x1],num[x2]....num[xk]这k个数都比l~r区间剩下的(下标不是x1...xk)的任何一个数大.题目就是给m个这种信息然后构造一个符合条件的数列 知道了这一点可以发现每一个信息都是一组偏序关系,即num[x1] > l~r区间剩下的数 .....num[…

Count Color Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 41202   Accepted: 12458 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.…