3109. [CQOI2013]新数独【DFS】

Description

Input

输入一共15行，包含一个新数独的实例。第奇数行包含左右方向的符号（<和>），第偶数行包含上下方向的符号（^和v）。

 

Output

输出包含9行，每行9个1~9的数字，以单个空格隔开。输入保证解惟一。

Sample Input

< > > < > <
v v ^ ^ v v ^ ^ ^
< < > < > <
^ ^ ^ v ^ ^ ^ v v
< < < < > >
> < > > > >
v ^ ^ ^ ^ v v v ^
> > > > < >
v v ^ v ^ v ^ v ^
> < < > > >
< < < < > <
v ^ v v v v ^ ^ v
< > > < < >
^ v v v ^ v ^ v v
< > < > < >

Sample Output

4 9 1 7 3 6 5 2 8
2 3 7 8 1 5 6 4 9
5 6 8 2 4 9 7 3 1
9 1 3 6 5 4 8 7 2
8 5 4 9 7 2 1 6 3
7 2 6 3 8 1 9 5 4
3 4 9 5 6 8 2 1 7
1 8 5 4 2 7 3 9 6
6 7 2 1 9 3 4 8 5

比靶形数独那个题水多了= =……

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #define id(x,y) (x-1)*9+y
 using namespace std;
 int Relation[][],ans[][],maxn;
 int dx[]= {,,-},dy[]= {,-,};
 bool row[][],column[][],square[][][],flag;
 char st[];

 void Dfs(int x,int y)
 {
     maxn=max(x,maxn);
     if (x== && y==)
     {
         flag=true;
         for (int i=; i<=; ++i)
         {
             for (int j=; j<=; ++j)
                 printf("%d ",ans[i][j]);
             printf("%d\n",ans[i][]);
         }
         return;
     }
     int down=,up=;
     for (int i=; i<=; ++i)
     {
         int xx=x+dx[i],yy=y+dy[i];
         if (xx< || xx> || yy< || yy> || ans[xx][yy]==) continue;
         if ( Relation[id(x,y)][id(xx,yy)] ==  ) down=max(down,ans[xx][yy]+);
         if ( Relation[id(x,y)][id(xx,yy)] ==  ) up=min(up,ans[xx][yy]-);
     }
     for (int i=down; i<=up; ++i)
         if (!row[x][i] && !column[y][i] && !square[(x-)/][(y-)/][i])
         {
             ans[x][y]=i;
             row[x][i]=column[y][i]=square[(x-)/][(y-)/][i]=true;
             if (y==) Dfs(x+,);
             else Dfs(x,y+);
             ans[x][y]=;
             row[x][i]=column[y][i]=square[(x-)/][(y-)/][i]=false;

             if (flag) return;
         }
 }

 int main()
 {
     memset(Relation,-,sizeof(Relation));
     for (int i=; i<=; ++i)
         if (i%==(^(i>= && i<=)))
         {
             for (int j=; j<=; ++j)
                 for (int k=; k<=; ++k)
                 {
                     char opt=getchar();
                     while (opt!='>' && opt!='<') opt=getchar();
                     if (opt=='>')
                     {
                         Relation[(j-)*+k+(i/+(i>))*][(j-)*+k+(i/+(i>))*+]=;
                         Relation[(j-)*+k+(i/+(i>))*+][(j-)*+k+(i/+(i>))*]=;
                     }
                     else
                     {
                         Relation[(j-)*+k+(i/+(i>))*+][(j-)*+k+(i/+(i>))*]=;
                         Relation[(j-)*+k+(i/+(i>))*][(j-)*+k+(i/+(i>))*+]=;
                     }
                 }

         }
         else
         {
             for (int j=; j<=; ++j)
             {
                 char opt=getchar();
                 while (opt!='^' && opt!='v') opt=getchar();
                 if (opt=='v')
                 {
                     Relation[id(i/+(i>=),j)][id(i/+(i>=)+,j)]=;
                     Relation[id(i/+(i>=)+,j)][id(i/+(i>=),j)]=;
                 }
                 else
                 {
                     Relation[id(i/+(i>=)+,j)][id(i/+(i>=),j)]=;
                     Relation[id(i/+(i>=),j)][id(i/+(i>=)+,j)]=;
                 }
             }
         }
     Dfs(,);
 }