CodeForces 1281 D. Beingawesomeism

Meaning of the title is not too long to copy over.

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Meaning of the questions is very simple is to $P$ all become $A$， $A$ can go in one direction and then traveled have become $A$ , is the biggest answer is clear and easy to get $4$ .
Then a few special cases judged on the line.
Specific comments in the code inside.

AC Code:

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a)
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

int n, t, m;
int res, ans, cnt, temp;
char s[100][100];
int main()
{
    sd(t);
    getchar();
    while (t--)
    {
        sdd(n, m);
        getchar();
        cnt = 0;
        ans = 4; //最多为4步
        rep(i, 1, n)
        {
            rep(j, 1, m)
            {
                sc(s[i][j]);
                if (s[i][j] == 'A')
                    cnt++;
            }
            getchar();
        }
        if (cnt == 0)
        {
            printf("MORTAL\n");
            continue;
        }
        else if (cnt == n * m)
        {
            printf("0\n");
            continue;
        }
        if (s[1][1] == 'A' || s[1][m] == 'A' || s[n][m] == 'A' || s[n][1] == 'A')
            ans = 2;
        rep(i, 1, n)
        {
            if (s[i][1] == 'A' || s[i][m] == 'A')
                ans = min(ans, 3);
        }
        rep(j, 1, m)
        {
            if (s[1][j] == 'A' || s[n][j] == 'A')
                ans = min(ans, 3);
        }
        rep(i, 1, n)
        {
            bool flag = 1;
            rep(j, 1, m)
            {
                if (s[i][j] == 'P')
                    flag = 0;
            }
            if (flag)
                ans = min(ans, 2 - (i == 1 || i == n));
        }
        rep(j, 1, m)
        {
            bool flag = 1;
            rep(i, 1, n)
            {
                if (s[i][j] == 'P')
                    flag = 0;
            }
            if (flag == 1)
                ans = min(ans, 2 - (j == 1 || j == m));
        } //如果这一行(列)全是A的情况，如果这一行(列)是第一行(列)或者最后一行(列)答案就是2，否则就是3
        pd(ans);
    }
    return 0;
}