PAT 甲级 1034 Head of a Gang (30 分)(bfs,map,强连通)
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 between the two persons. A "Gang" is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threthold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.
Input Specification:
Each input file contains one test case. For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:
Name1 Name2 Time
where Name1
and Name2
are the names of people at the two ends of the call, and Time
is the length of the call. A name is a string of three capital letters chosen from A
-Z
. A time length is a positive integer which is no more than 1000 minutes.
Output Specification:
For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.
Sample Input 1:
8 59
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 1:
2
AAA 3
GGG 3
Sample Input 2:
8 70
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 2:
0
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的头的字典序排序。
题解:
bfs求连通块。有几个注意点:
1、给定的是n条记录,n<=1000, 这说明人数应该是2000以内而不是1000以内。否则会测试点3运行时错误。
2、每一条记录都要考虑,bfs时不能仅判断这个节点是否被访问,还应该设置边的vis数组。
错误点:
1、忘记排序,测试点2,5答案错误,否则会测试点3运行时错误
2、n<=1000, 这说明人数应该是2000以内而不是1000以内
AC代码:
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<string>
#include<cstring>
using namespace std;
map<string,int>m1;
map<int,string>m2;
vector<int>v[];//给定的是n条记录,n<=1000, 这说明人数应该是2000以内而不是1000以内。测试点3运行时错误
int ok[];//标记有没有被访问过
struct node{//存储最终答案
string head;
int num;
}ans[];
int w[];//计算每个点的权重
int n,minw;
queue<int>q;
int k=;//map里的个数
int ans_num=;//答案个数
bool cmp(node x,node y){
return x.head<y.head;
}
void bfs(int x){
int w_sum=;//计算总权重
int num=;//计算人数
int max_w=w[x];
int head=x;
q.push(x);
ok[x]=;//标记有没有被访问过
num++;
w_sum+=w[x];
while(!q.empty()){
int a=q.front();
q.pop();
for(int i=;i<v[a].size();i++){
int b=v[a].at(i);
if(ok[b]) continue;
if(max_w<w[b]){//找出权重的为头目
head=b;
max_w=w[b];
}
ok[b]=;
q.push(b);
w_sum+=w[b];
num++;
}
}
w_sum/=;
if(num>= && w_sum>minw){
//cout<<"答案:num"<<num<<" "<<m2[head]<<" "<<ans_num<<" w_sum "<<w_sum<<endl;
node answer;
answer.num=num;
answer.head=m2[head];
ans[++ans_num]=answer;
}
}
int main(){
cin>>n>>minw;
memset(w,,sizeof(w));
memset(ok,,sizeof(ok));
for(int i=;i<=n;i++){
if(!v[i].empty()) v[i].clear();
}
for(int i=;i<=n;i++){
string na1,na2;
int ww;
cin>>na1>>na2>>ww;
if(!m1[na1]) {
m2[k]=na1;
m1[na1]=k++;
}
if(!m1[na2]) {
m2[k]=na2;
m1[na2]=k++;
}
w[m1[na1]]+=ww;
w[m1[na2]]+=ww;
v[m1[na1]].push_back(m1[na2]);
v[m1[na2]].push_back(m1[na1]);
}
for(int i=;i<=n;i++){
if(!ok[i]){
bfs(i);
}
}
cout<<ans_num<<endl;
sort(ans+,ans++ans_num,cmp);//排序,一开始忘记排了
for(int i=;i<=ans_num;i++){
cout<<ans[i].head<<" "<<ans[i].num<<endl;
}
return ;
}
PAT 甲级 1034 Head of a Gang (30 分)(bfs,map,强连通)的更多相关文章
- pat 甲级 1034. Head of a Gang (30)
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 wa ...
- PAT甲级1034. Head of a Gang
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. ...
- pat 甲级 1034 ( Head of a Gang )
1034 Head of a Gang (30 分) One way that the police finds the head of a gang is to check people's pho ...
- PAT 甲级 1064 Complete Binary Search Tree (30 分)(不会做,重点复习,模拟中序遍历)
1064 Complete Binary Search Tree (30 分) A Binary Search Tree (BST) is recursively defined as a bin ...
- PAT 甲级 1053 Path of Equal Weight (30 分)(dfs,vector内元素排序,有一小坑点)
1053 Path of Equal Weight (30 分) Given a non-empty tree with root R, and with weight Wi assigne ...
- PAT 甲级 1038 Recover the Smallest Number (30 分)(思维题,贪心)
1038 Recover the Smallest Number (30 分) Given a collection of number segments, you are supposed to ...
- 【PAT甲级】1034 Head of a Gang (30 分)
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 ...
- PAT Advanced 1034 Head of a Gang (30) [图的遍历,BFS,DFS,并查集]
题目 One way that the police finds the head of a gang is to check people's phone calls. If there is a ...
- PAT甲级1034 Head of a Gang【bfs】
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 ...
随机推荐
- 前端学习笔记--CSS样式--背景和超链接
1.背景 2.超链接: 举例:
- Mybatis-Generator逆向工程,简单策略
1.下载generator包 https://github.com/mybatis/generator/releases mybatis-generator-core-1.3.6.zip 官网下载即可 ...
- Java实习生面试题分享
1.Java有那些基本数据类型,String是不是基本数据类型,他们有何区别. Java语言提供了八种基本类型: 六种数字类型(四个整数型,两个浮点型) 字节型byte 8位 短整型short 16位 ...
- 02_View
1.View 1.基于类的视图 Class-based Views REST framework提供APIView是Django的View的子类 发送到View的Request请求:是REST fra ...
- emit传多个参数
https://blog.csdn.net/lxy123456780/article/details/87811113 子组件: this.$emit('closeChange',false,true ...
- element ui table的所有属性
1. table 的props: data: { type: Array, default: function() { return []; } }, size: String, width: [St ...
- 提高 github.com 项目下载速度
1 注册一个 github.com 账号 2 进入你感兴趣的项目 3 Fork 一个副本到你的账号之下 4 git clone https://github.com/your-name/fork-p ...
- Neo4j 在Linux下的安装登录
第一步:安装JDK https://blog.csdn.net/qq_33951308/article/details/82933535 第二步:下载并安装neo4j 下载地址 或者直接用wget ...
- 后缀数组 TYVJ P1860 后缀数组
/*P1860 后缀数组时间: 1000ms / 空间: 131072KiB / Java类名: Main描述 我们定义一个字符串的后缀suffix(i)表示从s[i]到s[length(s)]这段子 ...
- 论自动AC机
O(∩_∩)O哈哈~第一篇原创博客.终于结束了我“无敌转载王”的称号了!!!好开心! (⊙v⊙)嗯,看到标题觉得我是神犇的人,请再次仔细看看标题,是“自动AC”,而非“AC自动”哦!这是利用lemon ...