总结:能这么DP就这么写!
多练位运算标记。

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

const int N=55;
const int INF=0x3f3f3f3f;

int n, m, dp[N][2][2][2];
char s[N][N];

bool isdigit(char ch)
{
    if(ch>='0' && ch<='9')
        return true;
    return false;
}

bool isletter(char ch)
{
    if(ch>='a' && ch<='z')
        return true;
    return false;
}

bool issymbols(char ch)
{
    if(ch=='*' || ch == '&' || ch == '#')
        return true;
    return false;
}

int solve(int pos, bool digit, bool letter, bool symbols)
{
    if(pos==n)
    {
        if(digit && letter && symbols)
            return 0;
        return INF;
    }
    if(dp[pos][digit][letter][symbols] != -1)
        return dp[pos][digit][letter][symbols];
    int ans=INF;
    for(int j=0; j<m; j++)
    {
        int cost = min(j, m-j);
        if(isdigit(s[pos][j]))
            ans = min(ans, cost+ solve(pos+1, 1, letter, symbols));
        else if(isletter(s[pos][j]))
            ans = min(ans, cost+ solve(pos+1, digit, 1, symbols));
        else if(issymbols(s[pos][j]))
            ans = min(ans, cost+ solve(pos+1, digit, letter, 1));
    }
    return dp[pos][digit][letter][symbols]=ans;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0; i<n; i++)
        scanf("%s",s[i]);
    memset(dp, -1, sizeof(dp));
    printf("%d\n",solve(0,0,0,0));
    return 0;
}