leetcode Sudoku Solver
Write a program to solve a Sudoku puzzle by filling the empty cells.
Empty cells are indicated by the character '.'.
You may assume that there will be only one unique solution.
A sudoku puzzle...
...and its solution numbers marked in red.
32ms
Solution {
public:
struct Point
{
int x,y;
Point(int _x,int _y):x(_x),y(_y){}
Point(){}
};
vector v;
bool h[10][10];
bool c[10][10];
bool ge[10][10];
bool dfs(int cur,vector > &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 > &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);
}
}; 方法二：DLX精确覆盖（估计只有acmer才会用）
4ms
#define MAXM 10
#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 > &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, 81 + (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, 81 + (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++;
}
}
}
};