【BFS+hash】The Social Network

The Social Network

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1795    Accepted Submission(s): 555

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
-
-
  

 

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

概括题意：找不是自己朋友的朋友的朋友。

1：用map给每个姓名一个数字映射。

2：姓名上限要开2*MAX，可能是每对姓名都不相同。

3：其他就是BFS的搜索的，尽可能优化吧。

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#include<string>
#include<cstring>
using namespace std;
#define MAX 2005
map<string,int> mat;//姓名数字映射
vector<int> num[MAX];//某人的朋友
string name[MAX],out[MAX];//name存姓名，out存输出的姓名
int countx[MAX];//计数
bool flag[MAX];//是否是查找人的朋友
int cnt,u,v;
int main()
{
    int T,N,Q;
    scanf("%d",&T);
    for(int ti=1;ti<=T;++ti)
    {
            char s1[25],s2[25];
            string si;
            cnt=0;
            scanf("%d%d",&N,&Q);
            for(int i=0;i<N;++i)
                num[i].clear();
            mat.clear();
            for(;N--;)
            {
                    scanf("%s %s",s1,s2);
                    si=string(s1);  
                    if(mat.count(si))
                          u=mat.find(si)->second;  
                    else  
                    {  
                          u=cnt;
                          name[cnt]=si;  
                          mat.insert(make_pair(si,cnt++));  
                    }  
                    si=string(s2);  
                    if(mat.count(si))
                          v=mat.find(si)->second;  
                    else  
                    {
                          v=cnt; 
                          name[cnt]=si;    
                          mat.insert(make_pair(si,cnt++));  
                    }
                    num[u].push_back(v);
                    num[v].push_back(u);
            }
            printf("Case %d:\n",ti);
            for(;Q--;)
            {
                    scanf("%s",s1);
                    si=string(s1);
                    if(!mat.count(si)) continue;
                    memset(countx,0,sizeof(countx));
                    memset(flag,false,sizeof(flag));
                    int tmp=mat.find(si)->second;
                    for(int i=0;i<num[tmp].size();++i)
                        flag[num[tmp][i]]=true;
                    for(int i=0;i<cnt;++i)
                        if(flag[i])
                                   for(int j=0;j<num[i].size();++j)
                                               countx[num[i][j]]++;
                    int maxx=0;
                    for(int i=0;i<cnt;++i)
                        if(countx[i]>maxx&&(!flag[i])&&(i!=tmp))
                           maxx=countx[i];
                    if(maxx==0)  
                    {   
                        printf("-\n");  
                        continue;  
                    }  
                    int tmt=0;
                    for(int i=0;i<cnt;++i)
                        if(countx[i]==maxx&&(!flag[i])&&(i!=tmp))
                           out[tmt++]=name[i];
                    sort(out,out+tmt);
                    for(int i=0;i<tmt;++i)
                    {
                            if(i!=0) printf(" ");
                            printf("%s",out[i].c_str());
                    }
                    printf("\n");
            }
    }
    return 0;
}

来源：http://blog.youkuaiyun.com/ACM_Ted