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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#768738#7733. Cool, It’s Yesterday Four Times MoreDLYdly1105WA 1ms5668kbC++143.3kb2024-11-21 14:07:542024-11-21 14:07:54

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你现在查看的是最新测评结果

  • [2024-11-21 14:07:54]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:5668kb
  • [2024-11-21 14:07:54]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
char s[N][N],t[N][N];
int T,n,m,tot,id[N][N],tot2,id2[N<<1][N<<1];
int fx[10][10],vis[N][N],ok[N][N<<2],ok2[N<<2];
int ans;
queue<pair<int,int>>q;
void bfs(int qx,int qy)
{
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)vis[i][j]=0;
    q.push(make_pair(qx,qy));
    vis[qx][qy]=1;
    ok[id[qx][qy]][id2[n][m]]=1;
    ok2[id2[n][m]]=1;
    while(!q.empty())        
    {
        int x=q.front().first,y=q.front().second;
        q.pop();
        for(int i=0;i<4;i++)
        {
            int nx=x+fx[i][0],ny=y+fx[i][1];
            if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&s[nx][ny]=='.'&&!vis[nx][ny])
            {
                ok[id[qx][qy]][id2[nx-qx+n][ny-qy+m]]=1;
                ok2[id2[nx-qx+n][ny-qy+m]]=1;
                vis[nx][ny]=1;
                q.push(make_pair(nx,ny));
            }
        }
    }
    return;
}
bitset<N*2>b[N],aa,bb;
void solve()
{
    for(int x=1;x<=n;x++)for(int y=1;y<=m;y++)
    {
        if(s[x][y]!='.')continue;
        b[id[x][y]]=0;
        for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
        {
            b[id[x][y]][(i-1)*m+j]=ok[id[x][y]][id2[i-x+n][j-y+m]];
        }
    }
    int fl,fl2;
    b[0]=1;
    for(int i=1;i<=n*m;i++)b[0][i]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(s[i][j]!='.')continue;
            fl2=1;
            for(int x=1;x<=n;x++)
            {
                for(int y=1;y<=m;y++)
                {
                    if(x==i&&j==y)continue;
                    if(s[x][y]!='.')continue;
                    fl=0;
                    for(int r=1+x-i;r<=n+x-i;r++)
                    {
                        int ll=max(1,1+y-j),rr=min(m,m+y-j),t=r-(x-i);
                        aa=((b[id[i][j]]>>((t-1)*m+1))&(b[0]>>(n*m-m+1)));
                        if(r<1||r>n)bb=0;
                        else bb=(((b[id[x][y]]>>((r-1)*m+ll))&(b[0]>>(n*m-(rr-ll))))<<(m-rr));
                        if((aa&bb)!=aa)
                        {
                            fl=1;
                            break;
                        }
                    }
                    if(!fl){fl2=0;break;}
                }
                if(!fl2)break;
            }
            if(fl2)ans++;
        }
    }
    return;
}
int main()
{
    freopen("a.in","r",stdin);
    // freopen("kangaroo.out","w",stdout);
    fx[0][0]=0,fx[0][1]=-1;
    fx[1][0]=0,fx[1][1]=1;
    fx[2][0]=-1,fx[2][1]=0;
    fx[3][0]=1,fx[3][1]=0;
    scanf("%d",&T);
    while(T--)
    {
        tot=tot2=ans=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%s",s[i]+1);
        if(n>m)
        {
            for(int i=1;i<=m;i++)for(int j=1;j<=n;j++)t[i][j]=s[j][i];
            swap(n,m);
            for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)s[i][j]=t[i][j];
        }
        for(int i=1;i<=2*n-1;i++)for(int j=1;j<=2*m-1;j++)id2[i][j]=++tot2;
        for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(s[i][j]=='.')id[i][j]=++tot;
        for(int i=1;i<=tot;i++)for(int j=1;j<=tot2;j++)ok[i][j]=0;
        for(int i=1;i<=tot2;i++)ok2[i]=0;
        for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(s[i][j]=='.')bfs(i,j);
        solve();
        printf("%d\n",ans);
    }
    return 0;
}
/*
2
2 4
OOO.
OO..
2 3
OOO
.O.
*/

Details

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Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 5668kb

input:

4
2 5
.OO..
O..O.
1 3
O.O
1 3
.O.
2 3
OOO
OOO

output:


result:

wrong answer 1st lines differ - expected: '3', found: ''