You are an all-powerful being and you have created a rectangular world. In fact, your world is so bland that it could be represented by a r×cr×c grid. Each cell on the grid represents a country. Each country has a dominant religion. There are only two religions in your world. One of the religions is called Beingawesomeism, who do good for the sake of being good. The other religion is called Pushingittoofarism, who do murders for the sake of being bad.
Oh, and you are actually not really all-powerful. You just have one power, which you can use infinitely many times! Your power involves missionary groups. When a missionary group of a certain country, say aa, passes by another country bb, they change the dominant religion of country bb to the dominant religion of country aa.
In particular, a single use of your power is this:
You choose a horizontal 1×x1×x subgrid or a vertical x×1x×1 subgrid. That value of xx is up to you;
You choose a direction dd. If you chose a horizontal subgrid, your choices will either be NORTH or SOUTH. If you choose a vertical subgrid, your choices will either be EAST or WEST;
You choose the number ss of steps;
You command each country in the subgrid to send a missionary group that will travel ss steps towards direction dd. In each step, they will visit (and in effect convert the dominant religion of) all ss countries they pass through, as detailed above.
The parameters xx, dd, ss must be chosen in such a way that any of the missionary groups won’t leave the grid.
The following image illustrates one possible single usage of your power. Here, A represents a country with dominant religion Beingawesomeism and P represents a country with dominant religion Pushingittoofarism. Here, we’ve chosen a 1×41×4 subgrid, the direction NORTH, and s=2s=2 steps.
You are a being which believes in free will, for the most part. However, you just really want to stop receiving murders that are attributed to your name. Hence, you decide to use your powers and try to make Beingawesomeism the dominant religion in every country.
What is the minimum number of usages of your power needed to convert everyone to Beingawesomeism?
With god, nothing is impossible. But maybe you’re not god? If it is impossible to make Beingawesomeism the dominant religion in all countries, you must also admit your mortality and say so.
Input
The first line of input contains a single integer tt (1≤t≤2⋅1041≤t≤2⋅104) denoting the number of test cases.
The first line of each test case contains two space-separated integers rr and cc denoting the dimensions of the grid (1≤r,c≤601≤r,c≤60). The next rr lines each contains cc characters describing the dominant religions in the countries. In particular, the jj-th character in the ii-th line describes the dominant religion in the country at the cell with row ii and column jj, where:
“A” means that the dominant religion is Beingawesomeism;
“P” means that the dominant religion is Pushingittoofarism.
It is guaranteed that the grid will only contain “A” or “P” characters. It is guaranteed that the sum of the r⋅cr⋅c in a single file is at most 3⋅1063⋅106.
Output
For each test case, output a single line containing the minimum number of usages of your power needed to convert everyone to Beingawesomeism, or the string “MORTAL” (without quotes) if it is impossible to do so.
Example
Input
4
7 8
AAPAAAAA
PPPPAAAA
PPPPAAAA
APAAPPPP
APAPPAPP
AAAAPPAP
AAAAPPAA
6 5
AAAAA
AAAAA
AAPAA
AAPAP
AAAPP
AAAPP
4 4
PPPP
PPPP
PPPP
PPPP
3 4
PPPP
PAAP
PPPP
Output
2
1
MORTAL
4
Note
In the first test case, it can be done in two usages, as follows:
Usage 1:
Usage 2:
In the second test case, it can be done with just one usage of the power.
In the third test case, it is impossible to convert everyone to Beingawesomeism, so the answer is “MORTAL”.
思路：答案其实就6个，0,1,2,3,4,MORTAL。
①0的时候，全是A。
②MORTAL的时候，全是P。
③1的时候，在周围一圈任意一边全是A。
④2的时候，只要在四个角上有一个是A或者有一行或一列全是A。
⑤4的时候，在周围一圈没有A。
⑥其他情况就是3了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=61;
char s[maxx][maxx];
int n,m;
inline bool check1()
{
int flag1=1,flag2=1,flag3=1,flag4=1;
for(int i=0;i<m;i++) if(s[0][i]=='P') flag1=0;
for(int i=0;i<m;i++) if(s[n-1][i]=='P') flag2=0;
for(int i=0;i<n;i++) if(s[i][0]=='P') flag3=0;
for(int i=0;i<n;i++) if(s[i][m-1]=='P') flag4=0;
return flag1||flag2||flag3||flag4;
}
inline bool check2()
{
if(s[0][0]=='A'||s[0][m-1]=='A'||s[n-1][0]=='A'||s[n-1][m-1]=='A') return 1;
for(int i=0;i<n;i++)
{
int flag=1;
for(int j=0;j<m;j++) if(s[i][j]=='P') flag=0;
if(flag) return 1;
}
for(int j=0;j<m;j++)
{
int flag=1;
for(int i=0;i<n;i++) if(s[i][j]=='P') flag=0;
if(flag) return 1;
}
return 0;
}
inline bool check4()
{
int flag1=1,flag2=1,flag3=1,flag4=1;
for(int i=0;i<m;i++) if(s[0][i]=='A') flag1=0;
for(int i=0;i<m;i++) if(s[n-1][i]=='A') flag2=0;
for(int i=0;i<n;i++) if(s[i][0]=='A') flag3=0;
for(int i=0;i<n;i++) if(s[i][m-1]=='A') flag4=0;
return flag1&&flag2&&flag3&&flag4;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++) scanf("%s",s[i]);
int flag1=0,flag2=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(s[i][j]=='A') flag1=1;
else flag2=1;
}
if(flag1&&flag2) break;
}
if(flag2==0) cout<<0<<endl;
else if(flag1==0) cout<<"MORTAL"<<endl;
else
{
if(check1()) cout<<1<<endl;
else if(check2()) cout<<2<<endl;
else if(check4()) cout<<4<<endl;
else cout<<3<<endl;
}
}
return 0;
}
努力加油a啊，(o)/~
