「CF516B」Drazil and Tiles

2015年2月25日2,9710

Drazil created a following problem about putting 1 × 2 tiles into an n × m grid:

“There is a grid with some cells that are empty and some cells that are occupied. You should use 1 × 2 tiles to cover all empty cells and no two tiles should cover each other. And you should print a solution about how to do it.”

But Drazil doesn’t like to write special checking program for this task. His friend, Varda advised him: “how about asking contestant only to print the solution when it exists and it is unique? Otherwise contestant may print ‘Not unique‘ “.

Drazil found that the constraints for this task may be much larger than for the original task!

Can you solve this new problem?

Note that you should print ‘Not unique‘ either when there exists no solution or when there exists several different solutions for the original task.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2000).

The following n lines describe the grid rows. Character ‘.‘ denotes an empty cell, and the character ‘*‘ denotes a cell that is occupied.

Output

If there is no solution or the solution is not unique, you should print the string “Not unique“.

Otherwise you should print how to cover all empty cells with 1 × 2 tiles. Use characters “<>” to denote horizontal tiles and characters “^v” to denote vertical tiles. Refer to the sample test for the output format example.

Sample test(s)

input

3 3
...
.*.
...

3 3

...

.*.

...

output

Not unique

1	Not unique

input

4 4
..**
*...
*.**
....

4 4

..**

*...

*.**

....

output

&lt;&gt;**
*^&lt;&gt;
*v**
&lt;&gt;&lt;&gt;

<>**

*^<>

*v**

<><>

input

2 4
*..*
....

2 4

*..*

....

output

*&lt;&gt;*
&lt;&gt;&lt;&gt;

*<>*

<><>

input

1 1
.

1 1

output

Not unique

1	Not unique

input

1 1
*

1 1

output

*

Note

In the first case, there are indeed two solutions:

&lt;&gt;^
^*v
v&lt;&gt;

<>^

^*v

v<>

and

^&lt;&gt;
v*^
&lt;&gt;v

^<>

v*^

<>v

so the answer is “Not unique“.

题解

无脑爆搜会T。。。

正解拓扑排序。。。然后就不用说了吧。。

爆搜

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define ll long long 
#define inf 1000000000
#define dfsnxt(x,y) y==m?dfs(x+1,1):dfs(x,y+1)
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,flag;
char a[2005][2005],b[2005][2005];
bool s1(int i,int j)
{
    if(a[i+1][j]=='.'&&i+1<=n)
    {a[i][j]='^',a[i+1][j]='v';return 1;}
    return 0;
}
bool s2(int i,int j)
{
    if(a[i][j+1]=='.'&&j+1<=m)
    {a[i][j]='<',a[i][j+1]='>';return 1;}
    return 0;
}
void dfs(int x,int y)
{
    if(flag==2)return;
    if(x==n+1)
    {
        if(!flag)
        {
            for(int i=1;i<=n;i++)
                for(int j=1;j<=m;j++)
                    b[i][j]=a[i][j];
            flag=1;
        }
        else flag=2;
        return;
    }
    if(a[x][y]!='.')
    {
        dfsnxt(x,y);
        return;
    }
    if(s1(x,y))
    {
        dfsnxt(x,y);
        a[x][y]=a[x+1][y]='.';
    }
    if(s2(x,y))
    {
        dfsnxt(x,y);
        a[x][y]=a[x][y+1]='.';
    }
}
int main()
{
    n=read();m=read();
    for(int i=1;i<=n;i++)
        scanf("%s",a[i]+1);
    dfs(1,1);
    if(flag!=1)puts("Not unique");
    else 
        for(int i=1;i<=n;i++)
            printf("%s\n",b[i]+1);
    return 0;
}

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<algorithm>

#define ll long long

#define inf 1000000000

#define dfsnxt(x,y) y==m?dfs(x+1,1):dfs(x,y+1)

using namespace std;

int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,m,flag;

char a[2005][2005],b[2005][2005];

bool s1(int i,int j)

{

if(a[i+1][j]=='.'&&i+1<=n)

{a[i][j]='^',a[i+1][j]='v';return 1;}

return 0;

}

bool s2(int i,int j)

{

if(a[i][j+1]=='.'&&j+1<=m)

{a[i][j]='<',a[i][j+1]='>';return 1;}

return 0;

}

void dfs(int x,int y)

{

if(flag==2)return;

if(x==n+1)

{

if(!flag)

{

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

b[i][j]=a[i][j];

flag=1;

}

else flag=2;

return;

}

if(a[x][y]!='.')

{

dfsnxt(x,y);

return;

}

if(s1(x,y))

{

dfsnxt(x,y);

a[x][y]=a[x+1][y]='.';

}

if(s2(x,y))

{

dfsnxt(x,y);

a[x][y]=a[x][y+1]='.';

}

int main()

{

n=read();m=read();

for(int i=1;i<=n;i++)

scanf("%s",a[i]+1);

dfs(1,1);

if(flag!=1)puts("Not unique");

else

for(int i=1;i<=n;i++)

printf("%s\n",b[i]+1);

return 0;

}

拓扑排序

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define ll long long 
#define inf 1000000000
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
int d[2005][2005];
int qx[4000005],qy[4000005],top;
char ch[2005][2005];
int xx[4]={1,-1,0,0},yy[4]={0,0,1,-1};
void update(int nx,int ny)
{
    for(int i=0;i<4;i++)
    {
        int x=nx+xx[i],y=ny+yy[i];
        if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;
        d[x][y]--;
        if(d[x][y]==1)
        {           
            top++;
            qx[top]=x;qy[top]=y;
        }
    }
}
void solve()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(ch[i][j]=='.')
            for(int k=0;k<4;k++)
            {
                int x=i+xx[k],y=j+yy[k];
                if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;
                d[i][j]++;
            }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(d[i][j]==1)
            {
                top++;qx[top]=i;qy[top]=j;
            }
    while(top)
    {
        int nx=qx[top],ny=qy[top];top--;
        for(int k=0;k<4;k++)
        {
            int x=nx+xx[k],y=ny+yy[k];
            if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;
            if(k==0)ch[nx][ny]='^',ch[x][y]='v';
            if(k==1)ch[nx][ny]='v',ch[x][y]='^';
            if(k==2)ch[nx][ny]='<',ch[x][y]='>';
            if(k==3)ch[nx][ny]='>',ch[x][y]='<';
            d[nx][ny]=d[x][y]=0;
            update(x,y);
        }
    }
}
int main()
{
    n=read();m=read();
    for(int i=1;i<=n;i++)
        scanf("%s",ch[i]+1);
    solve();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(ch[i][j]=='.')
            {
                puts("Not unique");
                return 0;
            }
    for(int i=1;i<=n;i++)
        printf("%s\n",ch[i]+1);
    return 0;
}

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<algorithm>

#define ll long long

#define inf 1000000000

using namespace std;

int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

int n,m;

int d[2005][2005];

int qx[4000005],qy[4000005],top;

char ch[2005][2005];

int xx[4]={1,-1,0,0},yy[4]={0,0,1,-1};

void update(int nx,int ny)

{

for(int i=0;i<4;i++)

{

int x=nx+xx[i],y=ny+yy[i];

if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;

d[x][y]--;

if(d[x][y]==1)

{

top++;

qx[top]=x;qy[top]=y;

}

void solve()

{

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

if(ch[i][j]=='.')

for(int k=0;k<4;k++)

{

int x=i+xx[k],y=j+yy[k];

if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;

d[i][j]++;

}

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

if(d[i][j]==1)

{

top++;qx[top]=i;qy[top]=j;

}

while(top)

{

int nx=qx[top],ny=qy[top];top--;

for(int k=0;k<4;k++)

{

int x=nx+xx[k],y=ny+yy[k];

if(ch[x][y]!='.'||x<1||y<1||x>n||y>m)continue;

if(k==0)ch[nx][ny]='^',ch[x][y]='v';

if(k==1)ch[nx][ny]='v',ch[x][y]='^';

if(k==2)ch[nx][ny]='<',ch[x][y]='>';

if(k==3)ch[nx][ny]='>',ch[x][y]='<';

d[nx][ny]=d[x][y]=0;

update(x,y);

}

int main()

{

n=read();m=read();

for(int i=1;i<=n;i++)

scanf("%s",ch[i]+1);

solve();

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

if(ch[i][j]=='.')

{

puts("Not unique");

return 0;

}

for(int i=1;i<=n;i++)

printf("%s\n",ch[i]+1);

return 0;

}

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