Codeforce 593A 思维+枚举字符

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本文探讨了一种算法，用于从一系列字符串中选取满足特定条件的子集，目标是在不超过两种不同字符的前提下，最大化字符串总长度。通过遍历所有可能的字符组合，找到最优解。

A. 2Char

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

Copy

4
abb
cacc
aaa
bbb

output

Copy

input

Copy

5
a
a
bcbcb
cdecdecdecdecdecde
aaaa

output

Copy

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'}.

题目大意：给你n个字符串，从中选择可以多个字符串，前提条件是选择的字符串里面最多有连个不同的小写字母，求最长的这样的字符串的长度

这道题我被一组数据卡了半天，就是2 aaaa vvvv 我输出的是4，但是AC的答案应该是8，还是方法不对，看了题解才知道，原来可以枚举小写字母，每次枚举两个就可以了，因为题目要求的是最多有两个不同。具体看代码，算是学到一种骚操作了

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;i++)
int n;
char str[110][1010];
int main()
{
    int ans=0;
    int temp=0;
    scanf("%d",&n);
    rep(i,0,n-1)
    scanf("%s",str[i]);
    for(char i='a';i<='z';i++)
        for(char j=i+1;j<='z';j++)
    {
        temp=0;
        for(int k=0;k<n;k++)
        {
        int len1=strlen(str[k]);
        int l;
        for(l=0;l<len1;l++)
        if(str[k][l]!=i&&str[k][l]!=j) break;
        if(l==len1)
        temp+=len1;
        }
        ans=max(ans,temp);
    }
    printf("%d",ans);
}