codeforces 593A 2char

Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn’t written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.

Since the payment depends from the number of non-space characters in the article, Andrew wants to keep the words with the maximum total length.

Input

The first line of the input contains number n (1 ≤ n ≤ 100) — the number of words in the article chosen by Andrew. Following are n lines, each of them contains one word. All the words consist only of small English letters and their total length doesn’t exceed 1000. The words are not guaranteed to be distinct, in this case you are allowed to use a word in the article as many times as it appears in the input.

Output

Print a single integer — the maximum possible total length of words in Andrew’s article.

Sample test(s)

input

4

abb

cacc

aaa

bbb

output

9

input

5

a

a

bcbcb

cdecdecdecdecdecde

aaaa

output

6

Note

In the first sample the optimal way to choose words is {‘abb’, ‘aaa’, ‘bbb’}.

In the second sample the word ‘cdecdecdecdecdecde’ consists of three distinct letters, and thus cannot be used in the article. The optimal answer is {‘a’, ‘a’, ‘aaaa’}.

题目意思：只能用26个字母中的两个字母，在所给的n个串中找到能用两个字母组成的所有串最长总长度，所以若一个串含有三个以上字母，则抛弃掉；

#include <bits/stdc++.h>
using namespace std;

int main(){
ios::sync_with_stdio(false);

int n;
string s;
int k[26][26];  //记录只由字母a和字母b组成的所有串的总长度k[a][b];
while(cin >> n){

memset(k,0,sizeof(k)); // 初始化为0
while(n--) {
cin >> s;
char a = '0';  //定义临时变量字符a和字符b
char b = '0';
int flag = 1;
a = s[0]; //先让a为s[0];

//检查一个串中是否含有3个及以上不同字符
for(int i = 1; i < s.size(); i++) {
if(s[i] != a) {  //检查到一个不同的
if(b != s[i]) {   //看它是否等于b
if(b == '0') b = s[i];  //若此时b还未被赋值，则b赋值为s[i]，即串中已有两个不同的字符了
else {   //若b已经被赋值，则表示此时s[i] != a 且 != b,不满足条件，此串被标记为flag = 0；
flag = 0;
break;
}
}
}

}

if(flag) {   // flag 为1的串，即为未被抛弃的串
if(b == '0') k[a-'a'][a-'a'] = k[a-'a'][a-'a'] + s.size();  //如果未被赋值，则此串只由一个字符构成更新仅由此字符构成的串的长度k[a][a];
else k[a-'a'][b-'a'] = k[b-'a'][a-'a'] = k[a-'a'][b-'a'] + s.size();//否则，更新由a，b两个字符构成的串的长度
}
}

//  由k[a][b] = k[a][a] + k[b][b],更新所有由两个字符字符构成的串的总长度
for(int i = 0; i < 26; i++)
for(int j = 0; j < 26; j++) {
if(i == j ) continue;// 相同字符构成的串则不需要啦

k[i][j] = k[i][j] + k[i][i] + k[j][j];
}

int maxn = -1;
//查找最大值
for(int i = 0; i < 26; i++)
for(int j = 0; j < 26; j++){
if(maxn < k[i][j]) maxn = k[i][j];
}
cout << maxn << endl;

}
return 0;
}