[TS-A1488][2013中国国家集训队第二次作业]魔法波[高斯消元]

暴力直接解异或方程组，O(n^6)无法接受，那么我们考虑把格子分块，横着和竖着分别分为互不影响的块，这样因为障碍物最多不超过200个，那么块的个数最多为2*(800+200)=2000个，最后用bitset优化即可。

 #include <bits/stdc++.h>

 using namespace std;

 int n,cnt,Pos[],Right[],Left[];
 int Crs[][],Dwn[][],MID;
 char    str[][];

 bitset<>eq[];

 void    Gauss()
 {
     int i,j;
     for(i=;i<=cnt;++i)
     {
         j=i;
         while(!eq[j][i] && j<=cnt)j++;
         if(j==cnt+)continue;
         if(j!=i)swap(eq[j],eq[i]);
         for(j=;j<=cnt;++j)
             if(j!=i && eq[j][i])eq[j]^=eq[i];
     }
     return ;
 }

 int main()
 {
     int i,j,temp;

     scanf("%d",&n);

     for(i=;i<=n;++i)
         scanf("%s",str[i]+);

     for(i=;i<=n;++i)
         str[][i]=str[i][]=str[i][n+]=str[n+][i]='X';

     cnt=;

     for(j=;j<=n;++j) for(i=;i<=n+;++i)
     {
         if(str[i][j]!='X')Crs[i][j]=cnt;
         if(str[i-][j]=='X' && str[i][j]!='X')Left[cnt]=i,Pos[cnt]=j;
         if(str[i+][j]=='X' && str[i][j]!='X')Right[cnt]=i,cnt++;
     }

     MID=cnt-;

     for(i=;i<=n;++i) for(j=;j<=n+;++j)
     {
         if(str[i][j]!='X')Dwn[i][j]=cnt;
         if(str[i][j-]=='X' && str[i][j]!='X')Left[cnt]=j,Pos[cnt]=i;
         if(str[i][j+]=='X' && str[i][j]!='X')Right[cnt]=j,cnt++;
     }

     cnt--;

     for(i=;i<=MID;++i)
     {
         temp=;
         eq[i].set(i);

         for(j=Left[i];j<=Right[i];++j)
         {
             eq[i][Crs[j][Pos[i]]]=eq[i][Crs[j][Pos[i]]]^;
             eq[i][Dwn[j][Pos[i]]]=eq[i][Dwn[j][Pos[i]]]^;
             temp^=(str[j][Pos[i]]=='X'?:str[j][Pos[i]]-);
             temp^=;
         }

         eq[i][cnt+]=temp;
 #ifdef  Debug
         cerr << eq[i] << endl;
 #endif
     }

     for(i=MID+;i<=cnt;++i)
     {
         temp=;
         eq[i].set(i);

         for(j=Left[i];j<=Right[i];++j)
         {
             eq[i][Crs[Pos[i]][j]]=eq[i][Crs[Pos[i]][j]]^;
             eq[i][Dwn[Pos[i]][j]]=eq[i][Dwn[Pos[i]][j]]^;
             temp^=(str[Pos[i]][j]=='X'?:str[Pos[i]][j]-);
             temp^=;
         }

         eq[i][cnt+]=temp;
 #ifdef  Debug
         cerr << eq[i] << endl;
 #endif
     }

     Gauss();

 #ifdef  Debug
     cerr << endl;
     for(i=;i<=cnt;++i)
         cerr << eq[i] << endl;
     cerr << endl;
 #endif

     for(i=;i<=n;++i)
     {
         for(j=;j<=n;++j)
         {
             printf("%d",str[i][j]=='X'?:(^eq[Crs[i][j]][cnt+]^eq[Dwn[i][j]][cnt+]
                         ^(str[i][j]-)));
         }
         printf("\n");
     }

     return ;
 }