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…

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…

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…

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.…

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…

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…

题目如下: 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…

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…

分析: 考察并查集,注意中间合并时的时间的合并和人数的合并. #include <iostream> #include <stdio.h> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <cctype> #include <map> using namespace std; const i…

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…