This question is really difficult. Although it looks like a game theory problem, it can actually be solved using perfect matching.
All we need to do is find a region (connected) and color each cell with black and white. Then, if it is a perfect matching (bipartite graph), it means that no matter how the first player places their tiles, the second player can always place their tiles in a way that guarantees a win.
That is the perfect matching. However, if it is not a perfect matching, it means that there must be a position where the second player will definitely lose after placing their tile.
//============================================================================
// Name : Game of Tiles.cpp
// Date : 2013/3/18 下午2:17:15
// Author : GCA
//============================================================================
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 55
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
vector<int\> mz\[M\*M\],enable,benable;
int have\[M\*M\],bhave\[M\*M\];
int cmap\[M\]\[M\];
int nx,ny;
int color\[M\]\[M\];
int mx\[M\*M\],my\[M\*M\];
int d\[4\]\[2\]={{1,0},{-1,0},{0,1},{0,-1}};
void dfs\_color(int x,int y,int c,int depth){
for(int i=0;i<4;i++){
int newx=x+d\[i\]\[0\],newy=y+d\[i\]\[1\];
if(newx>=nx||newy>=ny||newx<0||newy<0||cmap\[newx\]\[newy\]=='X')continue;
int tmp=nx\*newy+newx;
if(c==0){
color\[x\]\[y\]=1;
mz\[nx\*y+x\].PB(tmp);
// mz\[tmp\].PB(nx\*y+x);
if(!have\[nx\*y+x\]){
enable.PB(nx\*y+x);
have\[nx\*y+x\]=1;
}
if(!bhave\[tmp\]){
benable.PB(tmp);
bhave\[tmp\]=1;
}
// debug("add %d %d %d\\n",nx\*y+x,tmp,depth);
if(color\[newx\]\[newy\])continue;
dfs\_color(newx,newy,1,depth+1);
}
else{
color\[x\]\[y\]=2;
if(color\[newx\]\[newy\])continue;
dfs\_color(newx,newy,0,depth+1);
}
}
}
int vis\[M\*M\];
bool dfs2(int x,int depth){
FOR(it,mz\[x\]){
int y=(\*it);
// if(x==63)printf("63y %d %d\\n",y,depth);
if(!vis\[y\]){
vis\[y\]=true;
if(my\[y\]==-1||dfs2(my\[y\],depth+1)){
mx\[x\]=y;
my\[y\]=x;
return true;
}
}
}
return false;
}
bool dfsAndB(int x,int y){
int cnt=0;
Set(have,0);
Set(bhave,0);
enable.clear();
benable.clear();
Set(mx,-1);
Set(my,-1);
enable.PB(nx\*y+x);
have\[nx\*y+x\]=1;
dfs\_color(x,y,0,0);
int numy=benable.size();
int numx=enable.size();
if(numx!=numy)return false;
FOR(it,enable){
Set(vis,0);
if(dfs2((\*it),0))cnt++;
}
// printf("numx %d numy %d cnt%d\\n",numx,numy,cnt);
// FOR(it,enable)printf("%d %d\\n",\*it,mx\[\*it\]);
return (cnt==numx);
}
bool solve(){
for(int i=0;i<ny;i++){
for(int j=0;j<nx;j++){
if(!color\[j\]\[i\]&&cmap\[j\]\[i\]!='X'){
// debug("=============\_\_\_\_\_\_\_===============\\n");
if(!dfsAndB(j,i))return true;
}
}
}
return false;
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d%d%\*c",&ny,&nx)){
Set(cmap,0);
Set(color,0);
for(int i=0;i<ny;i++){
for(int j=0;j<nx;j++){
cmap\[j\]\[i\]=getchar();
}
getchar();
}
if(solve())printf("1\\n");
else printf("2\\n");
for(int i=0;i<nx\*ny+1;i++)mz\[i\].clear();
}
}