761C Dasha and Password[简单dp]_considering that these are programming classes it -优快云博客

After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:

There is at least one digit in the string,
There is at least one lowercase (small) letter of the Latin alphabet in the string,
There is at least one of three listed symbols in the string: '#', '*', '&'.

$\text{[math]}$

Considering that these are programming classes it is not easy to write the password.

For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).

During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.

You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.

Input

The first line contains two integers n, m (3 ≤ n ≤ 50, 1 ≤ m ≤ 50) — the length of the password and the length of strings which are assigned to password symbols.

Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.

You have such input data that you can always get a valid password.

Output

Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.

   Examples 
 

     input 
   
3 4
1**2
a3*0
c4**

     output 
   
1

     input 
   
5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&

     output 
   
3

题意：给定m，n 接下来有m行n列的字符。每一行自成一个串，初始状态下，每行各有一个指针，指向每行的第一个字符，指针可移动，每次移动一个距离，可左右移动，（0位置左移动为n-1）求最小的移动次数，使得所有指针中至少包含一个小写字母，一个数字，一个*#&符号中的一个。

思路：首先对每一行字符串求得三个数值a，b，c，表示得到小写字母，得到数字，得到给定字符所用的最小次数。

然后对所有的abc遍历所有取得可能，只要保证不来自同一字符串即可。

代码：

#include<iostream>
#include<algorithm>
using namespace std;
int n, m;
int a[55], b[55], c[55];
char str[55];
int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; ++i)
        a[i] = b[i] = c[i] = 55;

    for(int i = 0; i < n; ++i)
    {
        cin >>str;
        for(int j = 0; j < m; ++j)
        {
            if(str[j] >= 'a' && str[j] <= 'z') a[i] = min(a[i], min(j, m - j));
            else if(str[j] >= '0' && str[j] <= '9') b[i] = min(b[i], min(j, m - j));
            else if(str[j] == '*' || str[j] == '#' || str[j] == '&') c[i] = min(c[i], min(j, m -j));
        }
    }

    int result = 0xffffff;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)  if(j != i)
                        for(int k = 0; k < n; ++k)  if(k != j)
                            result = min(result, a[i]+b[j]+c[k]);
    cout << result;
}