[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[2100],Right[2100],Left[2100];
int Crs[1100][1100],Dwn[1100][1100],MID;
char    str[1100][1100];

bitset<2100>eq[2100];

void    Gauss()
{
    int i,j;
    for(i=1;i<=cnt;++i)
    {
        j=i;
        while(!eq[j][i] && j<=cnt)j++;
        if(j==cnt+1)continue;
        if(j!=i)swap(eq[j],eq[i]);
        for(j=1;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=1;i<=n;++i)
        scanf("%s",str[i]+1);

    for(i=1;i<=n;++i)
        str[0][i]=str[i][0]=str[i][n+1]=str[n+1][i]='X';

    cnt=1;

    for(j=1;j<=n;++j) for(i=1;i<=n+1;++i)
    {
        if(str[i][j]!='X')Crs[i][j]=cnt;
        if(str[i-1][j]=='X' && str[i][j]!='X')Left[cnt]=i,Pos[cnt]=j;
        if(str[i+1][j]=='X' && str[i][j]!='X')Right[cnt]=i,cnt++;
    }

    MID=cnt-1;
    
    for(i=1;i<=n;++i) for(j=1;j<=n+1;++j)
    {
        if(str[i][j]!='X')Dwn[i][j]=cnt;
        if(str[i][j-1]=='X' && str[i][j]!='X')Left[cnt]=j,Pos[cnt]=i;
        if(str[i][j+1]=='X' && str[i][j]!='X')Right[cnt]=j,cnt++;
    }

    cnt--;

    for(i=1;i<=MID;++i)
    {
        temp=0;
        eq[i].set(i);

        for(j=Left[i];j<=Right[i];++j)
        {
            eq[i][Crs[j][Pos[i]]]=eq[i][Crs[j][Pos[i]]]^1;
            eq[i][Dwn[j][Pos[i]]]=eq[i][Dwn[j][Pos[i]]]^1;
            temp^=(str[j][Pos[i]]=='X'?0:str[j][Pos[i]]-48);
            temp^=1;
        }

        eq[i][cnt+1]=temp;
#ifdef  Debug
        cerr << eq[i] << endl;
#endif
    }

    for(i=MID+1;i<=cnt;++i)
    {
        temp=0;
        eq[i].set(i);

        for(j=Left[i];j<=Right[i];++j)
        {
            eq[i][Crs[Pos[i]][j]]=eq[i][Crs[Pos[i]][j]]^1;
            eq[i][Dwn[Pos[i]][j]]=eq[i][Dwn[Pos[i]][j]]^1;
            temp^=(str[Pos[i]][j]=='X'?0:str[Pos[i]][j]-48);
            temp^=1;
        }

        eq[i][cnt+1]=temp;
#ifdef  Debug
        cerr << eq[i] << endl;
#endif
    }

    Gauss();

#ifdef  Debug
    cerr << endl;
    for(i=1;i<=cnt;++i)
        cerr << eq[i] << endl;
    cerr << endl;
#endif

    for(i=1;i<=n;++i)
    {
        for(j=1;j<=n;++j)
        {
            printf("%d",str[i][j]=='X'?0:(1^eq[Crs[i][j]][cnt+1]^eq[Dwn[i][j]][cnt+1]
                        ^(str[i][j]-48)));
        }
        printf("
");
    }

    return 0;
}