简单模拟题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> using namespace std; struct X { int st; int len; int en; }p[]; queue<]; int n,m,k,s…

Waiting in Line Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of…

题面: 输入四个正整数N,M,K,Q(N<=20,M<=10,K,Q<=1000),N为银行窗口数量,M为黄线内最大人数,K为需要服务的人数,Q为查询次数.输入K个正整数,分别代表每位顾客需要被服务的时间,Q次查询每次查询一位顾客被服务完离开银行的时间.如果他在五点以前(不包括五点)没有开始被服务,输出“Sorry”. trick: 在五点以前(不包括五点)没有开始被服务,输出“Sorry”,而不是被服务完时间在五点以后(以题面为准). AAAAAccepted code: #inclu…

1014 Waiting in Line (30 分)   Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow li…

1014 Waiting in Line (30分)   Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow lin…

简单模拟题,注意读懂题意就行 #include <iostream> #include <queue> using namespace std; #define CUSTOMER_MAX 1000+1 #define INF 0x6fffffff #ifndef LOCAL // #define LOCAL #endif LOCAL int n; // number of windows <=20 int m ;// queue capacity <=10 int k;…

题目:https://www.patest.cn/contests/pat-a-practise/1014 思路: 直接模拟类的题. 线内的各个窗口各为一个队,线外的为一个,按时间模拟出队.入队. 注意点:即使到关门时间,已经在服务中的客户(窗口第一个,接待时间早于关门时间)还是可以被服务的.其它的则不服务. #include<iostream> #include<vector> #include<set> #include<map> #include<…

题意:n个窗口,每个窗口可以排m人.有k为顾客需要办理业务,给出了每个客户的办理业务时间.银行在8点开始服务,如果窗口都排满了,客户就得在黄线外等候.如果有一个窗口用户服务结束,黄线外的客户就进来一个.如果有多个可选,选窗口id最小的.输出查询客户的服务结束时间. 如果客户在17点(注意是包括的!!!就在这里被坑了,一开始还以为不包括...)或者以后还没开始服务,就输出Sorry如果已经开始了,无论多长都会继续服务的. 思路:建立一个优先级队列,存储在黄线之内的所有客户.对于m*n之前的人,依此…

这题写了有一点时间,最开始想着优化一下时间,用优先队列去做,但是发现有锅,因为忽略了队的长度. 然后思考过后,觉得用时间线来模拟最好做,先把窗口前的队列填满,这样保证了队列的长度是统一的,这样的话如果到某个时间,队首的人已经服务完了,这样这个队列的长度就减少一,这就变成了所有队列中长度最短的队列,所以直接向这个队列里面添加一个人就可以了. 按照从小到大轮询窗口的话,这样正好符合题中的要求,就是队列长度相同短,先选窗口序号小的窗口.如果按照这种策略,少一个就补一个的策略,队列长度会一直保持相等.…

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of each window is…

problem Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of each wi…

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of each window is…

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of each window is…

题目 Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are: The space inside the yellow line in front of each window…

利用广度优先搜索,找出每层的叶子节点的个数. #include <iostream> #include <vector> #include <queue> #include <fstream> using namespace std; vector<vector<int>> tree; vector<int> ans; void BFS(int s) { queue<pair<int, int>>…

模拟题.仔细一些即可. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<map> #include<queue> #include<cstring> #include<stack> #include<vector> #include<iostream> using namesp…

最短路. 每次询问的点当做起点,然后算一下点到其余点的最短路.然后统计一下最短路小于等于L的点有几个. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm>…

01背包路径输出. 保证字典序最小:从大到小做背包. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; +;…

最长公共子序列变形. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namespace std; int n,M,L; ],b[…

先找出可能在最短路上的边,图变成了一个DAG,然后在新图上DFS求答案就可以了. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<queue> #include<vector> using namespace std; const int INF=0x7FFFFFFF; ;…

1014. Waiting in Line (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for…

1014. Waiting in Line (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for…

PAT甲级1014. Waiting in Line 题意: 假设银行有N个窗口可以开放服务.窗前有一条黄线,将等候区分为两部分.客户要排队的规则是: 每个窗口前面的黄线内的空间足以包含与M个客户的一条线.因此,当所有N行都满时, 所有的客户(和包括)(NM + 1)第一个将不得不等待在黄线后面的一行. 每个客户将选择最短的行,以便在穿过黄线时等待.如果长度相同的两行以上,客户将始终选择数字最小的窗口. 客户[i]将在T [i]分钟内处理他/她的交易. 前N名客户被假定在上午8:00送达. 现在…

PAT (Advanced Level) Practice 1001-1005 PAT 计算机程序设计能力考试 甲级 练习题 题库:PTA拼题A官网 背景 这是浙大背景的一个计算机考试 刷刷题练练手 在博客更新题解 每五题更新一次 共155题 题目目录 1001 A+B Format (20分) 1002 A+B for Polynomials (25分) 1003 Emergency (25分) 1004 Counting Leaves (30分) 1005 Spell It Right (2…

PAT (Advanced Level) Practice 1042 Shuffling Machine (20 分) 凌宸1642 题目描述: Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employ…

PAT (Advanced Level) Practice 1006 Sign In and Sign Out (25 分) 凌宸1642 题目描述: At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of s…

Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性表:栈,队列,链表,顺序表 树:二叉树的建立,二叉树的遍历,完全二叉树,二叉查找树,平衡二叉树,堆,哈夫曼树 图:图的存储和遍历 经典高级算法: 深度优先搜索,广度优点搜索,回溯剪枝 贪心,并查集,哈希映射 最短路径(只考察过单源),拓扑排序(18年9月第一次涉及相关概念,未正式考过),关键路径(未…

PAT (Advanced Level) Practice 1046 Shortest Distance (20 分) 凌宸1642 题目描述: The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits. 译:你的任务很简单:给定 N 个出口,形成…

PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642 题目描述: Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10 4 ]. The first…

PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (o…