POJ 3074 Sudoku DLX精确覆盖

最新推荐文章于 2019-11-15 15:57:17 发布

转载最新推荐文章于 2019-11-15 15:57:17 发布 · 92 阅读

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原文链接：http://www.cnblogs.com/claireyuancy/p/6917322.html

本文介绍了一种使用DLX算法解决Sudoku难题的方法。DLX算法是一种回溯算法，特别适用于解决精确覆盖问题。文章详细展示了如何将Sudoku问题转化为精确覆盖问题，并通过DLX算法求解。

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DLX精确覆盖.....模版题

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8336		Accepted: 2945

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

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N=9;
const int maxn=N*N*N+10;
const int maxm=N*N*4+10;
const int maxnode=maxn*4+maxm+10;

char sudoku[maxn];

struct DLX
{
	int n,m,size;
	int U[maxnode],D[maxnode],L[maxnode],R[maxnode],Row[maxnode],Col[maxnode];
	int H[maxnode],S[maxnode];
	int ansd,ans[maxn];
	void init(int _n,int _m)
	{
		n=_n; m=_m;
		for(int i=0;i<=m;i++)
		{
			S[i]=0;
			U[i]=D[i]=i;
			L[i]=i-1;
			R[i]=i+1;
		}
		R[m]=0; L[0]=m;
		size=m;
		for(int i=1;i<=n;i++) H[i]=-1;
	}
	void Link(int r,int c)
	{
		++S[Col[++size]=c];
		Row[size]=r;
		D[size]=D[c];
		U[D[c]]=size;
		U[size]=c;
		D[c]=size;
		if(H[r]<0) H[r]=L[size]=R[size]=size;
		else
		{
			R[size]=R[H[r]];
			L[R[H[r]]]=size;
			L[size]=H[r];
			R[H[r]]=size;
		}
	}
	void remove(int c)
	{
		L[R[c]]=L[c]; R[L[c]]=R[c];
		for(int i=D[c];i!=c;i=D[i])
			for(int j=R[i];j!=i;j=R[j])
			{
				U[D[j]]=U[j];
				D[U[j]]=D[j];
				--S[Col[j]];
			}
	}
	void resume(int c)
	{
		for(int i=U[c];i!=c;i=U[i])
			for(int j=L[i];j!=i;j=L[j])
				++S[Col[U[D[j]]=D[U[j]]=j]];
		L[R[c]]=R[L[c]]=c;
	}
	bool Dance(int d)
	{
		if(R[0]==0)
		{
			for(int i=0;i<d;i++) sudoku[(ans[i]-1)/9]=(ans[i]-1)%9+'1';
			printf("%s\n",sudoku);
			return true;
		}
		int c=R[0];
		for(int i=R[0];i!=0;i=R[i])
			if(S[i]<S[c]) c=i;
		remove(c);
		for(int i=D[c];i!=c;i=D[i])
		{
			ans[d]=Row[i];
			for(int j=R[i];j!=i;j=R[j]) remove(Col[j]);
			if(Dance(d+1)) return true;
			for(int j=L[i];j!=i;j=L[j]) resume(Col[j]);
		}
		resume(c);
		return false;
	}
};

DLX dlx;

void place(int& r,int& c1,int& c2,int& c3,int& c4,int i,int j,int k)
{
	r=(i*N+j)*N+k;
	c1=i*N+j+1;
	c2=N*N+N*i+k;
	c3=N*N*2+N*j+k;
	c4=N*N*3+((i/3)*3+(j/3))*N+k;
}

int main()
{
	while(scanf("%s",sudoku)!=EOF)
	{
		if(sudoku[2]=='d') break;
		dlx.init(N*N*N,N*N*4);
		for(int i=0;i<N;i++)
		{
			for(int j=0;j<N;j++)
			{
				for(int k=1;k<=N;k++)
				{
					if(sudoku[i*N+j]=='.'||sudoku[i*N+j]==k+'0')
					{
						int r,c1,c2,c3,c4;
						place(r,c1,c2,c3,c4,i,j,k);
						dlx.Link(r,c1);
						dlx.Link(r,c2);
						dlx.Link(r,c3);
						dlx.Link(r,c4);
					}
				}
			}
		}
		dlx.Dance(0);
	}
	return 0;
}

转载于:https://www.cnblogs.com/claireyuancy/p/6917322.html