[POJ 2676] Sudoku【DLX】

  • 2018-01-27
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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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