字典树+AC自动机Hyper Prefix SetsＵＶＡ１１４８８YTACM第一周E


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

ACcode：

#include <iostream>
#include<cstdlib>
#include<cstdio>
#include<string.h>
#include<cstring>
using namespace std;
#define N 50001
#define M 202

int nn,ans;
int n,t;
struct DT//字典树结构体
{
int chid[2];
int num;//记录当前字符出现的次数
void init()
{
num=0;
memset(chid,-1,sizeof(chid));
}
} dt[N*10];
char str
[M];
void Init()//初始化
{
nn=ans=0;
for(int i=0; i<N*10; i++)
dt[i].init();
}

void Insert(char s[])//插入字符串到字典树中，并寻找最佳答案
{
int x=0;
int chnum;
int ls=strlen(s);
for(int i=0; i<ls; i++)
{
chnum=s[i]-'0';
if(dt[x].chid[chnum]==-1)
{
dt[x].chid[chnum]=++nn;
}
x=dt[x].chid[chnum];
dt[x].num++;
ans=max(ans,(i+1)*dt[x].num);//比较最大的goodness字符串前缀
}
}
void Read()//输入
{
for (int i=0; i<n; ++i)
{
scanf("%s", str[i]);
Insert(str[i]);
}
return;
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
Init();
Read();
printf("%d\n",ans);
}
}

Running Error是开的空间不够大