并查集解1335 : Email Merge(hihocoder)

题目

描述

You are given a list of usernames and their email addresses in the following format:

alice 2 [e-mail] [e-mail]

bob 1 [e-mail]

alicebest 2 [e-mail] [e-mail]

alice2016 1 [e-mail]

Your task is to merge the usernames if they share common email address:

alice alicebest alice2016

bob

输入

The first line contains an integer N, denoting the number of usernames. (1 < N <= 10000)

The following N lines contain N usernames and their emails in the previous mentioned format.

Each username may have 10 emails at most.

输出

Output one merged group per line.

In each group output the usernames in the same order as the input.

Output the groups in the same order as their first usernames appear in the input.

样例输入

4
alice 2 [e-mail] [e-mail]
bob 1 [e-mail]
alicebest 2 [e-mail] [e-mail]
alice2016 1 [e-mail]

样例输出

alice alicebest alice2016
bob

题目解析

题目给定n个人，每个人使用不多于10个的邮箱，要求找出共享邮箱的人（比如a使用[e-mail], b使用[e-mail]和[e-mail], c使用[e-mail]，则a，b，c共享邮箱）,要求找出这些类的人并按照输入顺序输出

思路解析

这就是并查集的简单使用样例，直接套入并查集的代码即可，注意要使用STL容器存储相关信息，以增快速率，比如用map存储邮箱以及该邮箱的使用者，这样能增加遍历速度，不然就可能出现TLE的情况

AC代码

//// Created by YoshiPark on 2017/10/16.//#include <iostream>#include <vector>#include <map>using namespace std;int F[10010];void init(int num) {    for (int i = 0; i < num; ++i) {        F[i] = i;    }}int getfather(int v) {    if (v == F[v]) {        return v;    } else{        F[v] = getfather(F[v]);        return F[v];    }}int merge(int a, int b) {    int fa = getfather(a);    int fb = getfather(b);    if (fa != fb) {        if (fa > fb) {            F[fa] = fb;        } else{            F[fb] = fa;        }    }}int main() {    int num;    cin >> num;    init(num);    map<string, vector<int>> maillist;    vector<string> userlist;    for (int i = 0; i < num; ++i) {        string username;        int mailnum;        cin >> username >> mailnum;        userlist.push_back(username);        for (int j = 0; j < mailnum; ++j) {            string mail;            cin >> mail;            maillist[mail].push_back(i);        }    }    for (auto it = maillist.begin(); it !=  maillist.end(); ++it) {        for (int j = 1; j < it->second.size(); ++j) {            merge(it->second[0],  it->second[j]);        }    }    vector<int> group[10010];    for(int i = 0; i< num; i++){        if(getfather(i) != i) group[getfather(i)].push_back(i);    }    for(int i = 0; i< num; i++){        if(getfather(i) == i){            cout<< userlist[i];            for(int j = 0; j<group[i].size(); j++){                cout<<" "<< userlist[group[i][j]];            }            cout<<endl;        }    }}