[POJ 2676] Sudoku【DLX】
Problem:
| Time Limit: 2000MS | Memory Limit: 65536K | Special Judge |
Description
Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task.
Input
The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.
Output
For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.
Sample Input
1 103000509 002109400 000704000 300502006 060000050 700803004 000401000 009205800 804000107
Sample Output
143628579 572139468 986754231 391542786 468917352 725863914 237481695 619275843 854396127
Source
Southeastern Europe 2005
Solution:
一道经典的搜索 DLX。
我们将数独转换为精确覆盖模型：
- 9 行 9 列中每格有且只有一个数 ⇒ 第 1 ~ 81 列
- 9 行中每行有且只有各一个 1 ~ 9 ⇒ 第 82 ~ 162 列
- 9 列中每列有且只有各一个 1 ~ 9 ⇒ 第 163 ~ 243 列
- 9 宫中每宫有且只有各一个 1 ~ 9 ⇒ 第 244 ~ 324 列
对于每个已确定的格子，插入 1 行，共 4 列，分别对应上述四个条件。
否则插入 9 行，每行 4 列，对应四个条件中分别填入 1 ~ 9 的情况。
用 Dancing Links 对这个至多 729 行 324 列 2916 个点的矩阵求精确覆盖即可。
详解可参考：http://blog.csdn.net/bl0ss0m/article/details/17918705。
这种方法一眨眼就能解出世界最难数独哦 ~~ ^O^ ~~
还有，POJ 这道题数据实在太弱了，写错了交上去还是 AC 的。。
Code: O(玄学) [828K, 0MS]
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
const int pal[10][10] = {
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
{0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
{0, 1, 1, 1, 2, 2, 2, 3, 3, 3},
{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
{0, 4, 4, 4, 5, 5, 5, 6, 6, 6},
{0, 7, 7, 7, 8, 8, 8, 9, 9, 9},
{0, 7, 7, 7, 8, 8, 8, 9, 9, 9},
{0, 7, 7, 7, 8, 8, 8, 9, 9, 9}
};
inline int getgrid(){
char ch;
while(!isdigit(ch = getchar()));
return ch - '0';
}
int T, sdk[10][10];
int col[5];
struct Dancing_Link{
int S[5000], C[5000];
int U[5000], D[5000], L[5000], R[5000];
int topv;
int x[5000], y[5000], v[5000];
inline void init(const int &col_num){
topv = col_num;
memset(S, 0, sizeof(S));
for(register int i = 0; i <= col_num; i++)
C[i] = U[i] = D[i] = i, L[i] = i - 1, R[i] = i + 1;
L[0] = col_num, R[col_num] = 0;
}
inline void del(const int &c){
L[R[c]] = L[c], R[L[c]] = R[c];
for(register int i = D[c]; i != c; i = D[i])
for(register int j = R[i]; j != i; j = R[j])
U[D[j]] = U[j], D[U[j]] = D[j], S[C[j]]--;
}
inline void add(const int &c){
L[R[c]] = R[L[c]] = c;
for(register int i = U[c]; i != c; i = U[i])
for(register int j = L[i]; j != i; j = L[j])
U[D[j]] = D[U[j]] = j, S[C[j]]++;
}
inline void addline(int *col, const int &col_size, const int &_row, const int &_col, const int &_val){
if(!col_size) return; // There may be empty lines
for(register int i = 1; i <= col_size; i++){
int &c = col[i];
C[++topv] = c;
L[topv] = topv - 1, R[topv] = topv + 1;
U[topv] = U[c], D[U[c]] = topv;
D[topv] = c, U[c] = topv;
S[c]++;
x[topv] = _row, y[topv] = _col, v[topv] = _val; // Record the original information
}
L[topv - col_size + 1] = topv, R[topv] = topv - col_size + 1;
}
inline void addstatus(const int &_row, const int &_col, const int &_val){
col[1] = (_row - 1) * 9 + _col;
col[2] = 81 + (_row - 1) * 9 + _val;
col[3] = 162 + (_col - 1) * 9 + _val;
col[4] = 243 + (pal[_row][_col] - 1) * 9 + _val;
addline(col, 4, _row, _col, _val);
}
inline bool dance(){
if(R[0] == 0) return 1;
int c = R[0];
for(register int i = R[c]; i; i = R[i])
if(S[i] < S[c]) c = i;
del(c);
for(register int i = D[c]; i != c; i = D[i]){
sdk[x[i]][y[i]] = v[i];
for(register int j = R[i]; j != i; j = R[j]) del(C[j]);
if(dance()) return 1;
for(register int j = L[i]; j != i; j = L[j]) add(C[j]);
}
add(c);
return 0;
}
} dl;
int main(){
scanf("%d", &T);
while(T--){
dl.init(324);
for(register int i = 1; i <= 9; i++)
for(register int j = 1; j <= 9; j++){
sdk[i][j] = getgrid();
if(sdk[i][j]) dl.addstatus(i, j, sdk[i][j]);
else for(register int k = 1; k <= 9; k++) dl.addstatus(i, j, k);
}
if(!dl.dance()) continue;
for(register int i = 1; i <= 9; i++){
for(register int j = 1; j <= 9; j++)
printf("%d", sdk[i][j]);
puts("");
}
}
return 0;
}
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