Joint StacksCrawling in process... Crawling failed Time Limit:4000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5818 uDebug Description   Input   Output   Sample Input   Sample Output   Hint   Description A…

Joint Stacks Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1819    Accepted Submission(s): 819 Problem Description A stack is a data structure in which all insertions and deletions of entries…

HDU  5818 Problem Description A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted is the first one that will be removed. In another wor…

Joint Stacks 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5818 Description A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted is th…

题目链接: Joint Stacks Time Limit: 8000/4000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the…

HDU 5818 Joint Stacks(联合栈) Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of…

Joint Stacks Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 828    Accepted Submission(s): 403 Problem Description A stack is a data structure in which all insertions and deletions of entries a…

多校7 HDU5818 Joint Stacks 题意:n次操作.模拟栈的操作,合并的以后,每个栈里的元素以入栈顺序排列 思路:开三个栈,并且用到了merge函数 O(n)的复杂度 #include <bits/stdc++.h> using namespace std; #define LL long long const int inf = 0x3f3f3f3f; ; ; #define clc(a,b) memset(a,b,sizeof(a)) ; void fre() {freope…

http://acm.hdu.edu.cn/showproblem.php?pid=5818 Joint Stacks Problem Description   A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted i…

题意:邀请k个朋友,每个朋友带有礼物价值不一,m次开门,每次开门让一定人数p(如果门外人数少于p,全都进去)进来,当最后所有人都到了还会再开一次门,让还没进来的人进来,每次都是礼物价值高的人先进.最后给出q个数,表示要输出第ni个进来的人的名字. 析:其实这就是一个模拟题,很容易知道是优先队列模拟,不能set,会超时,反正我是超时了,然后就一步步的模拟就好了,注意它给的时间可能不是按顺序,要排序,在比赛时我就没排序,一直WA.... 代码如下: #include <cstdio> #inclu…

Alisha 举办聚会,会在一定朋友到达时打开门,并允许相应数量的朋友进入,带的礼物价值大的先进,最后一个人到达之后放外面的所有人进来.用优先队列模拟即可.需要定义朋友结构体,存储每个人的到达顺序以及携带礼品价值,并重载<运算符.然后模拟即可. #include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<algorithm> #include…

Joint Stacks                                                                       Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                                    …

Alisha’s Party Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 4389    Accepted Submission(s): 1121 Problem Description Princess Alisha invites her friends to come to her birthday party. Each…

优先队列模拟一下就好. #include<bits/stdc++.h> using namespace std; priority_queue<int>q; int main() { int n; scanf("%d",&n); int t; scanf("%d",&t); ; i <= n; i++){ int v; scanf("%d",&v); q.push(v); } ; while(…

题意:有两个栈A和B,有3种操作:push,pop,merge.前两种都是栈的操作,最后一种表示的是如果“merge A B”,那么把B中的元素全部放到A中,且满足先入后出的栈原则. 分析:显然,我们给每一个节点配备一个时间戳即可.我一开始的思路是直接开两个优先队列进行直接模拟,merge操作就是把一个栈的元素全部倾倒到另一个栈内,但是会出现的问题是,如果有一个状态A和B的元素都相当多了,并且反复的进行merge操作,那么每一次merge都意味着大量的元素进出,肯定会超时的.因此,我们需要优化,…

题目:这里 题意: 两个类似于栈的列表,栈a和栈b,n个操作,push a x表示把数x放进a栈的栈底,pop b 表示将栈b的栈顶元素取出输出,并释放这个栈顶元素,merge a b表示把后面的那个 栈里的元素全部压进前面的那个栈里面,并且压入后前面的栈的所有元素按其进栈顺序排列然后后面的栈变为了空. push和pop操作都还好,就是优先队列或者栈的操作,主要是merge操作,如果啊按照题目来模拟每次将后面的栈元素取出放入前面的栈,可能会超时,而超时的主要因素是每次都将 元素多的栈压入了元素少…

A stack is a data structure in which all insertions and deletions of entries are made at one end, called the "top" of the stack. The last entry which is inserted is the first one that will be removed. In another word, the operations perform in a…

Description Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed h and the processor smashes k centimeters of potato each second. If there are less than k centimeters remai…

https://vjudge.net/problem/HDU-5818 题意:给你两个栈AB,有常规push,pop操作,以及一个merge操作,merge A B 即将A.B的元素按照入栈顺序全部出栈并推入栈A(merge B A 即反) 题解:用一个C来辅助,每一次merge将AB中的数据依次压入C中(如何依次?用vector<pair<int,int> >,second 存入栈的id,先按id弹出到tmp,再压入c),然后清空.之后A B若pop到底则从C pop(题目保证C…

用两个栈模拟,并保存每个点的时间戳.每次合并的时候记录合并时的时间戳mcnt和此时的topa和topb记做ta.tb. 每次pop的时候,如果栈的top的时间戳大于mcnt,则普通地pop,否则就在两个栈ta和tb下面找时间戳最大且还没pop掉的.然后用bj[时间戳]来标记已经pop了. #include <cstdio> #include <cstring> #define N 100005 using namespace std; struct node{ int id,v;…

原题链接 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In…

F - Mishka and trip Sample Output   Hint In the first sample test: In Peter's first test, there's only one cycle with 1 vertex. First player cannot make a move and loses. In his second test, there's one cycle with 1 vertex and one with 2. No one can…

http://codeforces.com/contest/723/problem/C 题目是给出一个序列 a[i]表示第i个歌曲是第a[i]个人演唱,现在选出前m个人,记b[j]表示第j个人演唱歌曲的数量, 现在你可以改变a[]这个数组,使得前m个人的演唱歌曲的数量的最小值最大. 很明显对于前m个人,每个人唱了多少首是很容易统计出来的,只需要对<=m的人进行统计即可,不用统计>m的,这样的话数组只需开到2000即可. 对于后面的人,可以改变,优先改变前m个人中演唱歌曲最小的那个,这个可以用优…

http://www.gdutcode.sinaapp.com/problem.php?cid=1049&pid=4 Problem E: 穷游中国在统题 Description Travel_poorly队是广工大目前最年轻的金牌队伍,队内成员分别是tmk.YFQ.Maple.这天他们接到教练的order要给新生杯统题,统题是个不轻松的工作,要评估各个题目的难度,设计出一套有梯度的套题,使得做题的情况有区分度.tmk很快想出了解决的办法,他给每道题目都设置了一个难度值,然后按照难度值进行筛选题…

1216: [HNOI2003]操作系统 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 1172  Solved: 649[Submit][Status][Discuss] Description 写一个程序来模拟操作系统的进程调度.假设该系统只有一个CPU,每一个进程的到达时间,执行时间和运行优先级都是已知的.其中运行优先级用自然数表示,数字越大,则优先级越高.如果一个进程到达的时候CPU是空闲的,则它会一直占用CPU直到该进程结束.除非在这个…

数据规模: 用优先队列对各个事件的发生先后记录即可: #include<iostream> #include<queue> using namespace std; ]; struct node { int no; int begin; int end; int type;//表示借,1表示时在还 node(int no, int begin,int end, int type):no(no),begin(begin),end(end),type(type) { } friend…

Alisha’s Party Problem's Link Mean: Alisha过生日,有k个朋友来参加聚会,由于空间有限,Alisha每次开门只能让p个人进来,而且带的礼物价值越高就越先进入. 给出开门的操作:当第ti个人来时,打开门让pi个人进入. q个询问,对于每个询问qi,你需要回答第qi个进入房间的人的名字. analyse: 维护一个优先队列,模拟就行. 需要注意,由于priority_queue对重载运算符貌似会出错,所以使用multiset来做priority_queue使用…

比赛的时候虽然考虑到没门的情况,但是写了几组都能过,就没想了,23333,差一行代码就能A,遗憾~~ #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #define MOD 100000000…

Alisha’s Party Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2650    Accepted Submission(s): 722 Problem Description Princess Alisha invites her friends to come to her birthday party. Each o…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5818 [题目大意] 给出两个栈A B(初始时为空),有三种操作: push.pop.merge. 其中merge是按照A B中元素进栈的相对顺序来重排的. [题解] 我们在当A,B栈出现第一个元素时,我们以这个元素建立左偏树,将其root赋值给所属栈,对于插入和输出堆顶操作,我们直接利用左偏树的功能实现,至于A,B的合并,我们将两棵左偏树合并后将一个标志置空即可. [代码] #include <…