uva 11488 - Hyper Prefix Sets（Trie）

H	Hyper Prefix Sets

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 Output for Sample Input

4 4 0000 0001 10101 010 2 01010010101010101010 11010010101010101010 3 010101010101000010001010 010101010101000010001000 010101010101000010001010 5 01010101010100001010010010100101 01010101010100001010011010101010 00001010101010110101 0001010101011010101 00010101010101001

6 20 66 44

Problem Setter : Abdullah Al Mahmud

Special Thanks : Manzurur Rahman Kha

用Trie统计出每一个子串出现的次数，每个结点保存两个信息，这个结点代表的串的长度，和这个串出现的次数，最后扫描所有的节点，找到长度 * 次数的最大值。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 1000000 + 5;

struct Node{
    int sum,len;
}node[maxn];

int trie[maxn][2];

int sz;
int idx(char c){ return c - '0';}

void insert(char *s){
    int u = 0,len = strlen(s);
    for(int i = 0;i < len;i++){
        int c = idx(s[i]);
        if(!trie[u][c]){
            trie[u][c] = sz;
            node[sz].len = i+1;
            sz++;
        }
        u = trie[u][c];
        node[u].sum++;
    }
}

int main(){
    int t,n;
    char str[205];
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        memset(trie,0,sizeof(trie));
        for(int i = 0;i < maxn;i++) node[i].sum = 0;
        sz = 1;
        while(n--){
            scanf("%s",str);
            insert(str);
        }
        int ans = -1;
        for(int i = 1;i < sz;i++){
            ans = max(ans,node[i].len*node[i].sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}