36th Chengdu semi-finals Network Competition hdoj4039 The Social Network (graph creation + string processing), 36thhdoj4039

Source: Internet
Author: User

36th Chengdu semi-finals Network Competition hdoj4039 The Social Network (graph creation + string processing), 36thhdoj4039


This is a Chinese online competition in Chengdu one year.

This idea is very simple, but from the time point of consideration, it is best not to use matrix storage, I use the chain forward star.

Online query is used. Map is used for string numbers, because it is convenient. The friend to be recommended is actually a friend of a friend (this refers to a friend who is directly connected by a side in the figure ).

Therefore, you only need to search for friends and count them.

Note: there cannot be spaces in the output.

Code:

#include<iostream>using namespace std;#include<cstdio>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>int n,m;char s11[20],s22[20];string g[20100],l[20100];int next[201000],head[2010],key[201000];int num;void add(int u,int v){     key[num]=v;     next[num]=head[u];     head[u]=num++;}int main(){    int T,pp=0;    scanf("%d",&T);    while (T--)    {    map<string,int> hash;    int n,m,i,j,k;          string s1,s2;    int cnt=0;        num=0;    scanf("%d%d",&n,&m);    memset(head,-1,sizeof(head));        for (i=0;i<n;++i)        {        scanf("%s%s",s11,s22);        s1=string(s11);        s2=string(s22);        if (hash[s1]==0)           hash[s1]=++cnt,l[cnt]=s1;        if (hash[s2]==0)           hash[s2]=++cnt,l[cnt]=s2;                add(hash[s1],hash[s2]);        add(hash[s2],hash[s1]);        }    printf("Case %d:\n",++pp);                for (i=0;i<m;++i)        {        scanf("%s",s11);        s1=string(s11);        int p=hash[s1];        int f[20100],flag[20010];        memset(f,0,sizeof(f));        memset(flag,0,sizeof(flag));                for (k=head[p];k!=-1;k=next[k]) flag[key[k]]=-1;                for (k=head[p];k!=-1;k=next[k])              if (key[k]!=p)            {               for (j=head[key[k]];j!=-1;j=next[j])                   if (key[j]!=key[k] && key[j]!=p && flag[key[j]]==0)                   {                      f[key[j]]++;                   }            }        int Max=-1;        for (k=1;k<=cnt;++k)             {            Max=max(Max,f[k]);          //  printf("%d\n",f[k]);            }        if (Max==0)           {           printf("-\n");           continue;           }        int q=0;        for (k=1;k<=cnt;++k)            if (Max==f[k])               {               g[q++]=l[k];               }        sort(g,g+q);               for (k=0;k<q-1;++k)  cout << g[k] << " ";        cout << g[q-1];        cout << endl;        }            }    return 0;}




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