3109. [CQOI2013]新数独【DFS】
Description
.jpg)
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
< > < > < >
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= =……
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define id(x,y) (x-1)*9+y 5 using namespace std; 6 int Relation[101][101],ans[12][12],maxn; 7 int dx[4]= {0,0,-1},dy[4]= {0,-1,0}; 8 bool row[10][10],column[10][10],square[10][10][10],flag; 9 char st[101]; 10 11 void Dfs(int x,int y) 12 { 13 maxn=max(x,maxn); 14 if (x==10 && y==1) 15 { 16 flag=true; 17 for (int i=1; i<=9; ++i) 18 { 19 for (int j=1; j<=8; ++j) 20 printf("%d ",ans[i][j]); 21 printf("%d\n",ans[i][9]); 22 } 23 return; 24 } 25 int down=1,up=9; 26 for (int i=1; i<=2; ++i) 27 { 28 int xx=x+dx[i],yy=y+dy[i]; 29 if (xx<1 || xx>9 || yy<1 || yy>9 || ans[xx][yy]==0) continue; 30 if ( Relation[id(x,y)][id(xx,yy)] == 1 ) down=max(down,ans[xx][yy]+1); 31 if ( Relation[id(x,y)][id(xx,yy)] == 0 ) up=min(up,ans[xx][yy]-1); 32 } 33 for (int i=down; i<=up; ++i) 34 if (!row[x][i] && !column[y][i] && !square[(x-1)/3][(y-1)/3][i]) 35 { 36 ans[x][y]=i; 37 row[x][i]=column[y][i]=square[(x-1)/3][(y-1)/3][i]=true; 38 if (y==9) Dfs(x+1,1); 39 else Dfs(x,y+1); 40 ans[x][y]=0; 41 row[x][i]=column[y][i]=square[(x-1)/3][(y-1)/3][i]=false; 42 43 if (flag) return; 44 } 45 } 46 47 int main() 48 { 49 memset(Relation,-1,sizeof(Relation)); 50 for (int i=1; i<=15; ++i) 51 if (i%2==(1^(i>=6 && i<=10))) 52 { 53 for (int j=1; j<=3; ++j) 54 for (int k=1; k<=2; ++k) 55 { 56 char opt=getchar(); 57 while (opt!='>' && opt!='<') opt=getchar(); 58 if (opt=='>') 59 { 60 Relation[(j-1)*3+k+(i/2+(i>10))*9][(j-1)*3+k+(i/2+(i>10))*9+1]=1; 61 Relation[(j-1)*3+k+(i/2+(i>10))*9+1][(j-1)*3+k+(i/2+(i>10))*9]=0; 62 } 63 else 64 { 65 Relation[(j-1)*3+k+(i/2+(i>10))*9+1][(j-1)*3+k+(i/2+(i>10))*9]=1; 66 Relation[(j-1)*3+k+(i/2+(i>10))*9][(j-1)*3+k+(i/2+(i>10))*9+1]=0; 67 } 68 } 69 70 } 71 else 72 { 73 for (int j=1; j<=9; ++j) 74 { 75 char opt=getchar(); 76 while (opt!='^' && opt!='v') opt=getchar(); 77 if (opt=='v') 78 { 79 Relation[id(i/2+(i>=5),j)][id(i/2+(i>=5)+1,j)]=1; 80 Relation[id(i/2+(i>=5)+1,j)][id(i/2+(i>=5),j)]=0; 81 } 82 else 83 { 84 Relation[id(i/2+(i>=5)+1,j)][id(i/2+(i>=5),j)]=1; 85 Relation[id(i/2+(i>=5),j)][id(i/2+(i>=5)+1,j)]=0; 86 } 87 } 88 } 89 Dfs(1,1); 90 }
