Leetcode Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells is indicated by the character '.

Assume that there would be is only one unique solution.

A Sudoku Puzzle ...

... and its solution numbers marked in red.

Method One: Simple search (Incredibly, I was drunk too)

32ms

 Solution {public:struct Point {int x, y;
    Point (int _x,int _y): X (_x), Y (_y) {}-point () {}};
    Vector<point> v;
    BOOL H[10][10];
    BOOL C[10][10];
    BOOL GE[10][10];
        BOOL Dfs (int cur,vector<vector<char> > &AMP;MP) {if (cur >= v.size ()) return true;
        Point P = v[cur];
            for (int i = 1;i <= 9; i++) {if (!h[p.x][i] &&!c[p.y][i] &&!ge[p.x/3*3+p.y/3][i])
                {H[p.x][i] = c[p.y][i] = Ge[p.x/3*3+p.y/3][i] = 1;
                MP[P.X][P.Y] = char (i + ' 0 ');
                if (Dfs (CUR+1,MP)) return true;
            H[p.x][i] = C[p.y][i] = Ge[p.x/3*3+p.y/3][i] = 0;
    }} return false;
        } void Solvesudoku (Vector<vector<char> > &board) {memset (h,0,sizeof (h));
        Memset (C,0,sizeof (c));
        memset (Ge,0,sizeof (GE));
for (int i = 0;i < 9;i++) for (int j = 0;j < 9; j + +)        {if (board[i][j] = = '. ') V.push_back (Point (I,j));
                else {int x = board[i][j]-' 0 ';
                H[I][X] = 1;
                C[J][X] = 1;
            GE[I/3*3+J/3][X] = 1;
    }} dfs (0,board); }
};

Method Two: DLX accurate coverage (estimated only acmer will be used)

4ms

#define MAXM #define MAXL 324 #define MAXN 10000 #define INF 0x7FFFFFFF Char SD[MAXM][MAXM];
int L[MAXN], R[MAXN], U[MAXN], D[MAXN], H[MAXN];
int size, C[MAXN], S[MAXN], X[MAXN], Q[MAXN], VIS[MAXL * 3];
    void Init () {int i;
        for (i = 0; I <= MAXL; i++) {l[i + 1] = i;
        R[i] = i + 1;
        U[i] = d[i] = i;
    S[i] = 0;
	} R[maxl] = 0;
    memset (h,-1,sizeof (H));
Size = MAXL + 1;
    } void Remove (int c) {int i, J;
    L[R[C]] = l[c];
    R[L[C]] = r[c];
            for (i = d[c], i = c; i = D[i]) {for (j = r[i]; j = i; j = R[j]) {U[d[j]] = u[j];
            D[U[J]] = d[j];
        s[c[j]]--;
    }}} void Resume (int c) {int i, J;
    L[R[C]] = c;
    R[L[C]] = c;
            for (i = d[c], i = c; i = D[i]) {for (j = r[i]; j = i; j = R[j]) {U[d[j]] = j;
            D[U[J]] = j;
        s[c[j]]++; }}} inline void Link (int r, int c) {D[size] = D[c];
    U[size] = c;
    U[D[C]] = size;
    D[C] = size;
    if (H[r] < 0) H[r] = l[size] = r[size] = size;
        else {l[size] = H[r];
        R[size] = R[h[r]];
        L[R[H[R]] = size;
    R[H[R]] = size;
    } s[c]++;
    X[size] = r;
c[size++] = c;
    } bool Dance (int now) {int I, j, c, temp;
    if (r[0] = = 0) return true;
            for (temp = Inf,i = r[0]; i; i = R[i]) {if (S[i] < temp) {temp = s[i];
        c = i;
    }} Remove (c);
        for (i = d[c]; i = c; i = D[i]) {Vis[x[i]] = true;
        for (j = r[i]; J! = i; j = r[j]) Remove (C[j]);
        if (Dance (now + 1)) return true;
        for (j = l[i]; J! = i; j = l[j]) Resume (C[j]);
    Vis[x[i]] = false;
    } Resume (c);
return false;
       } class Solution {Public:void Solvesudoku (vector<vector<char> > &board) {Init (); for (int i = 1;i <= 9;
       i++) {for (int j = 1;j <= 9; j + +) Sd[i][j] = board[i-1][j-1];
         } int I, j, K, R; for (r = 0, i = 1; i < MAXM; i++) {for (j = 1; j < Maxm; J + +) {if
                (Sd[i][j] = = '. ')
						{for (k = 1; k < MAXM; k++) {//H[++r] =-1;
                        r++;
                        Q[r] = k;
                        Link (R, (i-1) * 9 + k);
                        Link (R, Bayi + (j-1) * 9 + k);
                        Link (R, 162 + ((i-1)/3 * 3 + (J-1)/3) * 9 + k);
                    Link (R, 243 + (i-1) * 9 + j);
					}} else {//h[++r] =-1;
                    r++;
                    K = sd[i][j]-' 0 ';
                    Q[r] = k;
                    Link (R, (i-1) * 9 + k);
                    Link (R, Bayi + (j-1) * 9 + k); Link (R, 162 + ((i-1)/3 * 3 + (J-1)/3) * 9 + k);
                Link (R, 243 + (i-1) * 9 + j);
        }}} memset (Vis, false, sizeof (VIS));
        Dance (0);
        int p = 0;
                for (i=1;i<=r;i++) {if (Vis[i]) {board[p/9][p%9] = char (q[i] + ' 0 ');
            p++; }
        }
    }
};