POJ3074-Sudoku

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10594		Accepted: 3813

Description

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

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

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

Source

Stanford Local 2006

题意：数独问题

解题思路：舞蹈链的精确覆盖

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cctype>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int maxn = 300005;

int n, m, a[20][20], tot, p[1000][3];
char ch[100];

struct DLX
{
int L[maxn], R[maxn], U[maxn], D[maxn];
int row[maxn], col[maxn], ans[maxn], sum[maxn];
int n, m, num, cnt;
void add(int k, int l, int r, int u, int d, int x, int y)
{
L[k] = l;   R[k] = r;   U[k] = u;
D[k] = d;   row[k] = x;  col[k] = 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);
sum[i] = 0;
}
L[0] = m, R[m] = 0, cnt = m + 1;
}
void insert(int x, int y)
{
int temp = cnt - 1;
if (row[temp] != x)
{
add(cnt, cnt, cnt, U[y], y, x, y);
U[D[cnt]] = cnt; D[U[cnt]] = cnt;
}
else
{
add(cnt, temp, R[temp], U[y], y, x, y);
R[L[cnt]] = cnt; L[R[cnt]] = cnt;
U[D[cnt]] = cnt; D[U[cnt]] = cnt;
}
sum[y]++, cnt++;
}
void remove(int k)
{
R[L[k]] = R[k], L[R[k]] = L[k];
for (int i = D[k]; i != k; i = D[i])
for (int j = R[i]; j != i; j = R[j])
{
D[U[j]] = D[j];
U[D[j]] = U[j];
sum[col[j]]--;
}
}
void resume(int k)
{
R[L[k]] = k, L[R[k]] = k;
for (int i = D[k]; i != k; i = D[i])
for (int j = R[i]; j != i; j = R[j])
{
D[U[j]] = j;
U[D[j]] = j;
sum[col[j]]++;
}
}
bool dfs(int k)
{
if (!R[0]) { num = k; return true; }
int now = R[0];
for (int i = now; i != 0; i = R[i])
if (sum[now] > sum[i]) now = i;
remove(now);
for (int i = D[now]; i != now; i = D[i])
{
ans[k] = row[i];
for (int j = R[i]; j != i; j = R[j]) remove(col[j]);
if (dfs(k + 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]);
printf("\n");
}
}dlx;

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);
x = (x - 1) / 3 * 3 + (y - 1) / 3 + 1;
dlx.insert(tot, 243 + 9 * (x - 1) + z);
}

int main()
{
while (~scanf("%s", ch + 1) && strcmp(ch + 1, "end"))
{
tot = 0;
dlx.reset(729, 324);
for (int i = 1; i <= 9; i++)
{
for (int j = 1; j <= 9; j++)
{
int temp = (i - 1) * 9 + j;
if (ch[temp] == '.') a[i][j] = 0;
else a[i][j] = ch[temp] - '0';
if (a[i][j]) insert(i, j, a[i][j]);
else
{
for (int k = 1; k <= 9; k++)
insert(i, j, k);
}
}
}
if (dlx.dfs(0)) dlx.display();
}
return 0;
}