Hyper Prefix Sets UVA - 11488 字典树裸题

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

题意:找出字符串集合的一个前缀，使得前缀长度*出现次数的值为最大。

方法:用字典树处理字符串，同时在树的节点处记录前缀出现的次数以及长度，最后求出最大值。

字典树的用法可以看这位大佬的博客，讲的很简单明了详细。

https://www.cnblogs.com/TheRoadToTheGold/p/6290732.html

顺便自己复习一下字典树。字典树的存储方法，用二维数组模拟树，trie[i][j]=k,代表节点号为i的第j个子节点节点号为k.我们在树上存储字符串时，在用边来代表字母，同时在边的末尾记录字符串的一些值。比如出现次数，字符串长度，以及单词结束标志等。

代码:

#include
using namespace std;
int trie[10000010][2];
int sum[10000010];
int deep[10000010];
int tot;
void Insert(char *s)
{
    int rt=0;
    int len=strlen(s);
    for(int i=0;i