HDU4039(map应用)

The Social Network

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2958 Accepted Submission(s): 870

Problem Description
The social network system (SNS) helps people to keep connecting with their friends. Every user in SNS has a friends list. The user can read news posted by the users in his friends list. Friend relation is symmetric - if A is a friend of B, B is always a friend of A.

Another important function in SNS is friend recommendation. One effective way to recommend friends is recommend by mutual friends. A mutual friend between two users A and B, is a user who is a friend of both A and B. A user can not be a friend of himself. For a specific user A, the system will recommend the user who is not himself or his friend, and has mutual friends with A. If more than one such user exists, recommend the one has most mutual friends with A. If still a tie exists, output all of them.

Input
The first line is a integer T (T≤100), the number of test case.
The beginning of each test case is two integers N and Q, the number of friend relationship and the number of query. 1 ≤ N, Q ≤ 1000
The following N lines each contain two different names separated by a single space. Each name consisted by only lowercase letters, and its length is less than or equal to 15. This means the two users are friends. No friend relationship will be given more than once.
The following Q lines each describe a query. Each line contain one user name. The data guarantee that this name appears at least once in above N lines.

Output
For each case, you should output one line containing “Case k: ” first, where k indicates the case number and counts from one. Then for each query, output one line, contains one or more names of recommended friends, separate by a single space, sorted by alphabetical order. If no persons can be recommended, output one line contains “-”.

Sample Input
1
10 11
hongshu digua
yingying hongshu
xmm hongshu
huaxianzi xmm
tangjiejie huaxianzi
xhmz yingying
digua xhmz
zt tangjiejie
xmm lcy
notonlysuccess ljq
hongshu
digua
yingying
xmm
huaxianzi
tangjiejie
xhmz
zt
lcy
notonlysuccess
ljq

Sample Output
Case 1:
xhmz
yingying
digua
digua tangjiejie yingying
hongshu lcy zt
xmm
hongshu
huaxianzi
hongshu huaxianzi
-
-

题意:给出n组朋友关系,有q个user推荐好友查询.每次按字典序输出user的推荐好友。推荐好友定义:①是user好友的好友.②不是user本身和user的好友

思路:map+暴搜

#include <iostream>
#include <string.h>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
const int MAXN=;
int n,m;
map<string,int> mp;
map<int,string> rp;
int tot;
int dep[MAXN];
vector<int> arc[MAXN];
vector<string> fri;
int vis[MAXN];
int getID(string name)
{
    if(mp[name]==)
    {
        mp[name]=tot;
        rp[tot]=name;
        tot++;
    }
    return mp[name];
}
void addedge(int u,int v)
{
    arc[u].push_back(v);
    arc[v].push_back(u);
}
int main()
{
    int T;
    cin>>T;
    for(int cas=;cas<=T;cas++)
    {
        mp.clear();
        rp.clear();
        for(int i=;i<tot;i++)
        {
            arc[i].clear();
        }
        tot=;
        cin>>n>>m;
        for(int i=;i<n;i++)
        {
            string A,B;
            cin>>A>>B;
            int u=getID(A);
            int v=getID(B);
            addedge(u,v);
        }
        cout<<"Case "<<cas<<":"<<endl;
        for(int i=;i<m;i++)
        {
            memset(dep,,sizeof(dep));
            memset(vis,,sizeof(vis));
            fri.clear();
            string name;
            cin>>name;
            int u=getID(name);
            for(int i=;i<arc[u].size();i++)
            {
                int v=arc[u][i];
                vis[v]=;//标记u的好友
            }
            for(int i=;i<arc[u].size();i++)
            {
                int v=arc[u][i];
                for(int j=;j<arc[v].size();j++)
                {
                    int to=arc[v][j];
                    if(to!=u&&!vis[to])    dep[to]++;//u的推荐好友不能是u自己和u的好友
                }
            }
            int mx=*max_element(dep,dep+tot);
            if(mx!=)
            {
                for(int i=;i<tot;i++)
                {
                    if(dep[i]==mx)
                    {
                        string name=rp[i];
                        fri.push_back(name);
                    }
                }
                sort(fri.begin(),fri.end());
                int len=fri.size();
                for(int i=;i<len-;i++)
                {
                    cout<<fri[i]<<" ";
                }
                cout<<fri[len-]<<endl;
            }
            else    cout<<"-"<<endl;
        }
    }
    return ;
}