题意 给出一排空花瓶 有两种操作  1是 从A花瓶开始放F朵花 如果当前瓶有花就跳过前往下一个 直到花用完或者 瓶子到了最后一个为止 输出 成功放花的第一个和最后一个  如果没有输出 can not....... 2是 清空 一段区间的花 并输出清空了多少朵花 思路:用线段树维护一段区间有多少空花瓶  1操作 就是两次二分 分别求起点和终点   起点  check条件是  query(1,a,mid)>0  终点是  query(1,二分出的起点,mid)>=min(f,query(1,a,n…

Vases and Flowers 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4614 Problem Description Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them…

题目传送门 题意:给出一些花开花落的时间,问某个时间花开的有几朵 分析:这题有好几种做法,正解应该是离散化坐标后用线段树成端更新和单点询问.还有排序后二分查找询问点之前总花开数和总花凋谢数,作差是当前花开的数量,放张图易理解: 还有一种做法用尺取法的思想,对暴力方法优化,对询问点排序后再扫描一遍,花开+1,花谢-1.详细看代码. 收获:一题收获很多:1. 降低复杂度可以用二分 2. 线段计数问题可以在端点标记1和-1 3. 离散化+线段树 终于会了:) (听说数据很水?) 代码1:离散化+线段树…

G - Queue HDU - 5493 题目大意:给你n个人的身高和这个人前面或者后面有多少个比他高的人,让你还原这个序列,按字典序输出. 题解: 首先按高度排序. 设每个人在其前面有k个人,设比这个人高有x个人,所以k=min(k,x-k),求出每一个人在他前面比他高的人的数量. 然后用线段树来维护一段区间的空位,因为对于每一个人来说比他高的人肯定都是出现在他后面,所以这个就说明它前面要空出几个空位. 但是线段树可以维护一段区间的空位,没办法准确找出我需要空位为一个准确的数的位置. 这个就可…

思路:当k为1的时候,用二分法查询包含有f个空瓶的上界r,然后更新会方便很多,直接更新区间(a,r)了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define lson(x) (x<<1) #define rson(x) ((x<<1)+1) #define mid ((tree[p…

Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flow…

Queue-jumpers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3348    Accepted Submission(s): 904 Problem Description Ponyo and Garfield are waiting outside the box-office for their favorite mo…

Transformation Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others) Total Submission(s): 4095    Accepted Submission(s): 1008 Problem Description Yuanfang is puzzled with the question below:  There are n integers, a1,…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4996    Accepted Submission(s): 1576 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

Level up Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3973    Accepted Submission(s): 1104 Problem Description Level up is the task of all online games. It's very boooooooooring. There is o…