POJ 3074 Sudoku

Description

In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,

Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.

Input

The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.

Output

For each test case, print a line representing the completed Sudoku puzzle.

Sample Input

.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534.
......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3.
end

Sample Output

527389416819426735436751829375692184194538267268174593643217958951843672782965341

416837529982465371735129468571298643293746185864351297647913852359682714128574936

建图+dlx

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=10005;
const ll size=15;
int n,m,x,y,T,a[size][size],tot,p[1000][3];
char s[200];

inline void read(int &ret)
{
  char c;
  do { c = getchar();
  } while(c < '0' || c > '9');
  ret = c - '0';
  while((c=getchar()) >= '0' && c <= '9')
  ret = ret * 10 + ( c - '0' );
}

struct DLX
{
  int L[maxn],R[maxn],U[maxn],D[maxn];
  int row[maxn],col[maxn],ans[maxn],cnt[maxn];
  int n,m,num,sz;
  void add(int now,int l,int r,int u,int d,int x,int y)
  {
    L[now]=l; R[now]=r; U[now]=u;
    D[now]=d;   row[now]=x;  col[now]=y;
  }
  void reset(int n,int m)
  {
    this->n=n; this->m=m;
    for (int i=0;i<=m;i++)
    {
      add(i,i-1,i+1,i,i,0,i);
      cnt[i]=0;
    }
    L[0]=m;   R[m]=0;   sz=m+1;
  }
  void insert(int x,int y)
  {
    int ft=sz-1;  
    if (row[ft]!=x)
    {
      add(sz,sz,sz,U[y],y,x,y);
      U[D[sz]]=sz; D[U[sz]]=sz;
    }
    else 
    {
      add(sz,ft,R[ft],U[y],y,x,y);
      R[L[sz]]=sz; L[R[sz]]=sz;
      U[D[sz]]=sz; D[U[sz]]=sz;
    }
    ++cnt[y]; ++sz;
  }
  void remove(int now)
  {
    R[L[now]]=R[now];
    L[R[now]]=L[now];
    for (int i=D[now];i!=now;i=D[i])
      for (int j=R[i];j!=i;j=R[j])
      {
        D[U[j]]=D[j];
        U[D[j]]=U[j];
        --cnt[col[j]];
      }
  }
  void resume(int now)
  { 
    for (int i=U[now];i!=now;i=U[i])
      for (int j=L[i];j!=i;j=L[j])
      {
        D[U[j]]=j;
        U[D[j]]=j;
        ++cnt[col[j]];
      }
    R[L[now]]=now;
    L[R[now]]=now;
  }
  bool dfs(int x)
  {
    if (!R[0]) {num=x; return true;}
    int now=R[0];
    for (int i=now;i!=0;i=R[i])
      if (cnt[now]>cnt[i]) now=i;
    remove(now);
    for (int i=D[now];i!=now;i=D[i])
    {
      ans[x]=row[i];
      for (int j=R[i];j!=i;j=R[j]) remove(col[j]);
      if (dfs(x+1)) return true;
      for (int j=L[i];j!=i;j=L[j]) resume(col[j]);
    }
    resume(now);
    return false;
  }
  void display()
  {
    for (int i=0;i<num;++i) 
    {
      a[p[ans[i]][0]][p[ans[i]][1]]=p[ans[i]][2];
    }
    for (int i=1;i<=9;i++)
      for (int j=1;j<=9;j++) printf("%d",a[i][j]);
    puts("");
  }
}dlx;

int check(int x,int y)
{
  x=(x-1)/3;
  y=(y-1)/3;
  return 3*x+y;
}

void insert(int x,int y,int z)
{
  p[++tot][0]=x;
  p[tot][1]=y;
  p[tot][2]=z;
  dlx.insert(tot,9*(x-1)+y);
  dlx.insert(tot,81+9*(x-1)+z);
  dlx.insert(tot,162+9*(y-1)+z);
  dlx.insert(tot,243+9*check(x,y)+z);
}

int main()
{
  int c[size][size],r[size][size],u[size][size];
  while (scanf("%s",s+1),strcmp("end",s+1))
  {
    tot=0;
    memset(c,0,sizeof(c));
    memset(r,0,sizeof(r));
    memset(u,0,sizeof(u));
    dlx.reset(729,324);
    for (int i=1;i<=9;++i)
      for (int j=1;j<=9;++j) 
      {
        if (s[(i-1)*9+j]=='.') a[i][j]=0;
        else a[i][j]=s[(i-1)*9+j]-'0';
        if (a[i][j]) 
        {
          insert(i,j,a[i][j]);
          c[i][a[i][j]]=r[j][a[i][j]]=u[check(i,j)][a[i][j]]=1;
        }
      }
    for (int i=1;i<=9;++i)
      for (int j=1;j<=9;++j) 
        for (int k=1;k<=9;++k)
          if (a[i][j]+c[i][k]+r[j][k]+u[check(i,j)][k]<1) insert(i,j,k);
    if (dlx.dfs(0)) dlx.display();
  }
  return 0;
}