Codeforces Round #394 (Div. 2)C. Dasha and Password【暴力】

C. Dasha and Password

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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: '#', '*', '&'.



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

Note
In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.



In the second test one of possible algorithms will be:

to move the pointer of the second symbol once to the right.
to move the pointer of the third symbol twice to the right.



题目大意：

这是一个类似密码锁的东西。

一开始的时候密码设置在第一列，我们可以通过拨动来改变每一行的密码。

合法的密码必须满足三个条件：

①至少有一个数字。

②至少有一个小写字母。

③至少有一个特殊字符（*&#）

问最少拨动多少下就能得到合法的密码。

思路：

因为只要满足这三个条件就能算是一个合法密码，那么也就是说，我们只要枚举三行使得这三行一个拨动到数字，一个拨动到小写字母，一个拨动到特殊字符即可。

这三行需要维护一个最小拨动次数。

那么我们的任务就是O（n^3）枚举这三行，接下来O(M)来维护当前情况的最小拨动次数即可

对于可行解，维护最小输出。

总时间复杂度O(n^3m);

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char a[500][800];
int b[50*50*50];
int c[50*50*50];
int d[50*50*50];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;i++)
{
scanf("%s",a[i]);
}
int output=0x3f3f3f3f;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
for(int k=1;k<=n;k++)
{
if(i==j||j==k||i==k)continue;
int sum1=0x3f3f3f3f;
for(int z=0;z<m;z++)
{
if(a[i][z]>='0'&&a[i][z]<='9')
{
sum1=min(sum1,min(z,m-z));
}
}
int sum2=0x3f3f3f3f;
for(int z=0;z<m;z++)
{
if(a[j][z]>='a'&&a[j][z]<='z')
{
sum2=min(sum2,min(z,m-z));
}
}
int sum3=0x3f3f3f3f;
for(int z=0;z<m;z++)
{
if(a[k][z]=='*'||a[k][z]=='#'||a[k][z]=='&')
{
sum3=min(sum3,min(z,m-z));
}
}
if(sum1==0x3f3f3f3f||sum2==0x3f3f3f3f||sum3==0x3f3f3f3f)continue;
output=min(output,sum1+sum2+sum3);
}
}
}
printf("%d\n",output);
}
}