UVA11488——Hyper Prefix Sets（字典树，最长前缀）

Prefix goodness of a set string is length of longest common prefix*number of strings in the set. For
example the prefix goodness of the set {000,001,0011} is 6.You are given a set of binary strings. Find
the maximum prefix 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 prefix 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

求最长公共前缀，第一个例子中，最长公共前缀是000，有两个这样的序列，所以是3*2=6

第二个例子中，最长公共前缀是0，所以可以把自己本身作为前缀，前缀的长度就是自己本身的长度，所以是20*1=20

用结构体数组实现字典树，child[0]表示该点之后有个0，里面存放的是这个0的位置

#include 
#include 
#include 
#define N 50010
#define M 210
using namespace std;
struct trie
{
    int child[2],shownum;
    void init()
    {
        shownum=0;
        memset(child,-1,sizeof(child));
    }
};
int n,nn,ant;
char str[N][M];
trie tr[10*N];
void init()
{
    for(int i=0;i<10*N;++i)
        tr[i].init();
}
void inset(char s[])
{
    int x=0,sn=strlen(s),num;
    for(int i=0;i