UV-11464 Even Parity

The first line of enumeration, in binary format, can be introduced in subsequent n-1 rows. Whether the push side verification meets the requirements. Save the optimal solution

 #include<cstdio>  #include<algorithm>  #include<iostream>  #include<cstring>  using namespace std;  int a[20][20];  int b[20][20];  int main()  {      int cas,n;      cin>>cas;      for(int ca=1;ca<=cas;ca++)      {          scanf("%d",&n);          memset(a,0,sizeof(a));          memset(b,0,sizeof(b));          for(int i=1;i<=n;i++)              for(int j=1;j<=n;j++)                  scanf("%d",&a[i][j]);          int top=(1<<n);          int ans=0x3f3f3f3f;          for(int f=0;f<top;f++)          {              int tmp=f;              int sum=0;              int ok=1;              for(int i=n;i>=1;i--)              {                  b[1][i]=(tmp&1);                  tmp>>=1;                  if(b[1][i]-a[1][i]<0) {ok=0;break;}                  if(b[1][i]!=a[1][i]) sum++;              }              if(!ok) continue;              for(int j=1;j<=n;j++)              {                  b[2][j]=((b[1][j-1]+b[1][j+1])&1);                  if(b[2][j]-a[2][j]<0) {ok=0;break;}                  if(a[2][j]!=b[2][j]) sum++;              }              if(!ok) continue;              for(int i=3;i<=n;i++)              {                  for(int j=1;j<=n;j++)                  {                      b[i][j]=((b[i-1][j-1]+b[i-1][j+1]+b[i-2][j])&1);                      if(b[i][j]-a[i][j]<0) {ok=0;break;}                      if(a[i][j]!=b[i][j]) sum++;                  }              }              if(!ok) continue;              for(int i=2;i<=n-1;i++)              {                  for(int j=1;j<=n;j++)                  {                      if(((b[i-1][j]+b[i+1][j]+b[i][j-1]+b[i][j+1])&1)) {ok=0;break;}                  }                  if(!ok) break;              }              if(!ok) continue;              for(int j=1;j<=n;j++)              {                  if((b[n][j-1]+b[n-1][j]+b[n][j+1])&1) {ok=0;break;}                  if(b[n][j]-a[n][j]<0) {ok=0;break;}              }              if(!ok) continue;  //            for(int i=1;i<=n;i++){  //                for(int j=1;j<=n;j++)  //                    cout<<b[i][j];cout<<endl;}  //            cout<<sum<<endl;              ans=min(ans,sum);          }          printf("Case %d: %d\n",ca,ans==0x3f3f3f3f?-1:ans);      }      return 0;  }