Purification - Red Huang

先判斷有沒有可能解
這題只要想成
XOO
這樣施放或是
XXX
OOO
如果有邪惡的區域就往下一格施放法術就好
這樣保證是最快且絕對可以全部消除
//====================================================================||  
// Name        : Purification.cpp                                              ||  
// Date : 2013/7/21 下午1:29:22                                               ||  
// Author : GCA                                                       ||  
//                  6AE7EE02212D47DAD26C32C0FE829006                  ||  
//====================================================================||  
#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 105  
#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);  
}  
int n;  
char mz\[M\]\[M\];  
int nx\[M\],ny\[M\];  
void solve(){  
    Set(nx,0);  
    Set(ny,0);  
    bool fx=true,fy=true;  
    for(int i=0;i<n;i++){  
        for(int j=0;j<n;j++){  
            if(mz\[i\]\[j\]=='E'){  
                nx\[i\]++;  
                ny\[j\]++;  
            }  
        }  
    }  
    for(int i=0;i<n;i++){  
        if(nx\[i\]==n){  
            fx=false;  
        }  
        if(ny\[i\]==n){  
            fy=false;  
        }  
    }  
    if(!fx&&!fy)puts("-1");  
    else{  
        if(fx){  
            for(int i=0;i<n;i++){  
                for(int j=0;j<n;j++){  
                    if(mz\[i\]\[j\]=='.'){  
                        printf("%d %d\\n",i+1,j+1);  
                        break;  
                    }  
                }  
            }  
        }else{  
            for(int i=0;i<n;i++){  
                for(int j=0;j<n;j++){  
                    if(mz\[j\]\[i\]=='.'){  
                        printf("%d %d\\n",j+1,i+1);  
                        break;  
                    }  
                }  
            }  
        }  
    }  
}  
int main() {  
    ios\_base::sync\_with\_stdio(0);  
    while(~scanf("%d",&n)){  
        for(int i=0;i<n;i++)scanf("%s",mz\[i\]);  
        solve();  
    }  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
}