Channel Allocation HDU1373

#include<bits/stdc++.h>
using namespace std;

#define N 27

int n;
int mp[N][N];
int ans;
int alt[N][N];
int Max[N];

bool dfs(int cur,int tot)//cur是s1集合的个数
{
    if(0==cur)
    {
        if(tot>ans)
        {
            ans=tot;return true;
        }
        return false;
    }

    for(int i=0;i<cur;i++)
    {
        if( tot+cur-i<=ans )return false;
        int u=alt[tot][i];
        if( Max[u]+tot<=ans )return false;
        int next=0;
        for(int j=i+1;j<cur;j++)
            if(mp[u][ alt[tot][j] ])alt[tot+1][next++]=alt[tot][j];
        if(dfs(next,tot+1)) return 1;
    }
    return 0;
}

int maxclique(void)
{
    ans=0;
    memset(Max,0,sizeof(Max));
    for(int i=n-1;i>=0;i--)
    {
        int cur=0;
        for(int j=i+1;j<n;j++)if(mp[i][j])alt[1][cur++]=j;//1为s1集合
          dfs(cur,1);
        Max[i]=ans;
    }
    return ans;
}

int main()
{   char s[30];
 
    while(scanf("%d",&n),n)
    {   
        memset(mp,0,sizeof (mp));
        for(int i=0;i<n;i++)
        {
          scanf("%s",s);
          for(int j=2;s[j];j++)
             mp[i][ s[j]-‘A‘   ]=mp[ s[j]-‘A‘   ][i]=1;
        }
        int ans=maxclique();
        if(ans==1) printf("1 channel needed.\n");
         else printf("%d channels needed.\n", ans);
    }
 return 0;
}