HDU 11488 hyper prefix sets (string-trie tree)

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 {000,001,001 1} 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 Khan

Question:

If a represents the length of a public prefix, and B represents the number of strings containing this prefix, ask you the maximum value of a * B.

Solution:

Create a trie tree, perform edge search, and directly update the maximum length multiplied by the number.

Solution code:

#include <iostream>#include <cstring>#include <string>#include <cstdio>using namespace std;const int maxn=500000;int tree[maxn][2];int val[maxn],cnt;int n,ans;void insert(string st){    int s=0;    for(int i=0;i<st.length();i++){        if( tree[s][st[i]-'0']==0 ) tree[s][st[i]-'0']=++cnt;        s=tree[s][st[i]-'0'];        val[s]++;        if((i+1)*val[s]>ans) ans=(i+1)*val[s];    }}void initial(){    cnt=ans=0;    memset(val,0,sizeof(val));    memset(tree,0,sizeof(tree));}void solve(){    cin>>n;    for(int i=0;i<n;i++){        string st;        cin>>st;        insert(st);    }    cout<<ans<<endl;}int main(){    int t;    cin>>t;    while(t-- >0){        initial();        solve();    }    return 0;}