Pre x goodness of a set string is length of longest common pre x*number of strings in the set. For
example the pre x goodness of the set f000,001,0011g is 6.You are given a set of binary strings. Find
the maximum pre x goodness among all possible subsets of these binary strings.
Input
First line of the input contains T ( 20) the number of test cases. Each of the test cases start with n
( 50000) the number of strings. Each of the next n lines contains a string containing only `0' and `1'.
Maximum length of each of these string is 200.
Output
For each test case output the maximum pre x goodness among all possible subsets of n binary strings.
Sample Input
4
4
0000
0001
10101
010
2
01010010101010101010
11010010101010101010
3
010101010101000010001010
010101010101000010001000
010101010101000010001010
5
01010101010100001010010010100101
01010101010100001010011010101010
00001010101010110101
0001010101011010101
00010101010101001
Sample Output
6
20
66
44
example the pre x goodness of the set f000,001,0011g is 6.You are given a set of binary strings. Find
the maximum pre x goodness among all possible subsets of these binary strings.
Input
First line of the input contains T ( 20) the number of test cases. Each of the test cases start with n
( 50000) the number of strings. Each of the next n lines contains a string containing only `0' and `1'.
Maximum length of each of these string is 200.
Output
For each test case output the maximum pre x goodness among all possible subsets of n binary strings.
Sample Input
4
4
0000
0001
10101
010
2
01010010101010101010
11010010101010101010
3
010101010101000010001010
010101010101000010001000
010101010101000010001010
5
01010101010100001010010010100101
01010101010100001010011010101010
00001010101010110101
0001010101011010101
00010101010101001
Sample Output
6
20
66
44
扔到一个Trie上,分别统计每个节点的答案
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<queue>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXNode (MAXN*MAXLen)
#define Sigma_size (2)
#define MAXN (50000+10)
#define MAXLen (200+10)
#define MAXT (20+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
class Trie
{
public:
int ch[MAXNode][Sigma_size];
int v[MAXNode],siz;
Trie(int _siz=0):siz(_siz){ans=0; MEM(ch) MEM(v)}
void mem(int _siz=0){siz=_siz; ans=0; MEM(ch) MEM(v) }
int idx(char c){return c-'0';}
void insert(char *s,int val=0)
{
int u=0,n=strlen(s);
Rep(i,n)
{
int c=idx(s[i]);
if (!ch[u][c])
{
++siz;
MEM(ch[siz]);
ch[u][c]=siz;
}
u=ch[u][c];
}
v[u]+=val;
}
void find(char *s)
{
int u=0,n=strlen(s);
Rep(i,n)
{
int c=idx(s[i]);
if (!ch[u][c])
{
return;
}
u=ch[u][c];
}
}
ll ans;
void calc(int u,ll l)
{
Rep(c,Sigma_size)
{
if (!ch[u][c]) continue;
calc(ch[u][c],l+1);
v[u]+=v[ch[u][c]];
}
ans=max(ans,v[u]*l);
}
}S;
int T,n;
char s[MAXLen];
int main()
{
// freopen("uva11488.in","r",stdin);
cin>>T;
while(scanf("%d",&n)==1)
{
S.mem();
For(i,n)
{
scanf("%s",s);
S.insert(s,1);
}
S.calc(0,0);
cout<<S.ans<<endl;
}
return 0;
}