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
|
题意:可以得到的权值为某个前缀的长度*这个前缀的字符串的个数。
思路:就是简单的Trie树,可以说是模板题了。
AC代码如下:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int ans;
char s[210];
struct Trie
{ int index;
Trie *next[10];
Trie()
{ index=0;
memset(next,0,sizeof(next));
}
};
void Trie_Insert(Trie *tr,char *s,int len)
{ tr->index++;
ans=max(ans,tr->index*len);
if(*s!='\0')
{ if(tr->next[*s-'0']==0)
tr->next[*s-'0']=new Trie;
Trie_Insert(tr->next[*s-'0'],s+1,len+1);
}
}
int main()
{ int t,n,i,j,k;
scanf("%d",&t);
while(t--)
{ Trie *root =new Trie;
scanf("%d",&n);
ans=0;
while(n--)
{ scanf("%s",s);
Trie_Insert(root,s,0);
}
printf("%d\n",ans);
}
}