用并查集分割团伙,判断输出~ #include<bits/stdc++.h> using namespace std; ; },weight[maxn]; unordered_map<string,int> pos1; unordered_map<int,string> pos2; unordered_map<string,int> ans; struct node { string id; ; ; }Node[maxn]; struct gang { st…

One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made b…

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…

Source: PAT A1034 Head of a Gang (30 分) Description: One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to b…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1034 此题考查并查集的应用,要熟悉在合并的时候存储信息: #include <iostream> #include <string> #include <map> #include <vector> #include <algorithm> #include <cstddef> using namespace std; struc…

PAT甲级1034. Head of a Gang 题意: 警方找到一个帮派的头的一种方式是检查人民的电话.如果A和B之间有电话,我们说A和B是相关的.关系的权重被定义为两人之间所有电话的总时间长度. "帮派"是超过2人的群体,彼此相关,总关系权重大于给定的阈值K.在每个帮派中,最大总重量的人是头.现在给了一个电话列表,你应该找到帮派和头. 输入规格: 每个输入文件包含一个测试用例. 对于每种情况,第一行分别包含两个正数N和K(均小于或等于1000),电话号码和权重.然后N行遵循以下格…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805456881434624 题意: 给定n条记录(注意不是n个人的记录),两个人之间的关系的权值为这两个人之间所有电话记录的时间之和. 一个连通块的权值为所有关系权值之和. 如果一个连通块节点数大于2,且权值大于给定的k,称这是一个gang,拥有关系权值和最多的人是gang的头. 要求输出gang的数量,每个gang的头,每个gang的人数.按照gang的头的字典…

One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made b…

给出n和k接下来n行,每行给出a,b,c,表示a和b之间的关系度,表明他们属于同一个帮派一个帮派由>2个人组成,且总关系度必须大于k.帮派的头目为帮派里关系度最高的人.(注意,这里关系度是看帮派里边的和,而不是帮派里所有个人的总和.如果是按个人算的话,相当于一条边加了两次,所以应该是>2*k)问你有多少个合格帮派,以及每个帮派里最大的头目是谁,按字典序输出 先并查集一下,然后统计每个帮的成员数.总关系度.以及头目 #include <iostream> #include <c…

1034 Head of a Gang (30)(30 分) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time lengt…

1034. Head of a Gang (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related.…

1034. Head of a Gang (30) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of…

1034 Head of a Gang (30 分)   One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length…

1034 Head of a Gang (30 分) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of…

题目 One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls mad…

昨天准备学完图相关的知识,但是学起来挺懵的,理解起来不难,但自己一回想,又什么都想不起来. 翻来覆去看图的遍历,还是觉得有点没到位. 所以做题来检测一下,果然学和自己做是两码事. 先看的书,又看的柳婼的代码.思路一样. 自己照着打了一遍,又自己实现了一遍,总体并不难,关键就是三十分的题,要花多点时间读懂题意. 只发现一个对于我来说的注意事项: 两个人打电话,可能不止打一次. #include<string> #include<iostream> #include<map>…

有一个两分的case出现段错误,真是没救了,估计是要写bfs的形式,可能栈溢出了 #include <cstdio> #include <cstdlib> #include <string> #include <vector> #include <unordered_map> #include <algorithm> using namespace std; ][] = {}; class Man { public: int id;…

简单DFS. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<string> #include<queue> #include<stack> #include<algorithm> #include<iostream> using namespace st…

题意: 输入两个正整数N和K(<=1000),接下来输入N行数据,每行包括两个人由三个大写字母组成的ID,以及两人通话的时间.输出团伙的个数(相互间通过电话的人数>=3),以及按照字典序输出团伙老大的ID和团伙的人数(团伙中通话时长最长的人视为老大,数据保证一个团伙仅有一名老大). AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string s…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

1034 Head of a Gang (30 分) One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between Aand B, we say that A and B is related. The weight of a relation is defined to be the total time length of…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1031. Hello World for U (20) 题意 将给定的字符串打印出 U 型. 比如给定helloworld,打印出 1 2 3 4 5 h d e l l r lowo 设定左边的字符个数为 n1,底边字符个数为 n2,右边字符个数为 n3.需要满足 n1 = n3…

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…

并查集 PAT (Advanced Level) Practice 并查集 相关题 <算法笔记> 重点摘要 1034 Head of a Gang (30) 1107 Social Clusters (30) 1118 Birds in Forest (25) <算法笔记> 9.6 并查集 重点摘要 1. 定义 father[i] 表示元素 i的父结点 father[i] = i 表示元素 i 为该集合根结点 每个集合只存在一个根结点,且其作为所属集合的标识 int father[…

Description: Telefraud(电信诈骗) remains a common and persistent problem in our society. In some cases, unsuspecting victims lose their entire life savings. To stop this crime, you are supposed to write a program to detect those suspects from a huge amou…

Telefraud(电信诈骗) remains a common and persistent problem in our society. In some cases, unsuspecting victims lose their entire life savings. To stop this crime, you are supposed to write a program to detect those suspects from a huge amount of phone c…

博客出处:http://blog.csdn.net/zhoufenqin/article/details/50497791 题目出处:https://www.patest.cn/contests/pat-a-practise 1001 题意:给出a+b,计算c=a+b,将c输出时每三个数加一个“,” 思路:数据范围比较小,根据特殊的数据范围,也可特殊求解,不解释 #include<iostream> #include<cstdio> #include<cstring>…