POJ 3076 Sudoku DLX精确覆盖

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DLX精确覆盖模版题.....

Sudoku Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 4416   Accepted: 2143 Description A Sudoku grid is a 16x16 grid of cells grouped in sixteen 4x4 squares, where some cells are filled with letters from A to P (the first 16 capital letters of the English alphabet), as shown in figure 1a. The game is to fill all the empty grid cells with letters from A to P such that each letter from the grid occurs once only in the line, the column, and the 4x4 square it occupies. The initial content of the grid satisfies the constraints mentioned above and guarantees a unique solution.    Write a Sudoku playing program that reads data sets from a text file. Input Each data set encodes a grid and contains 16 strings on 16 consecutive lines as shown in figure 2. The i-th string stands for the i-th line of the grid, is 16 characters long, and starts from the first position of the line. String characters are from the set {A,B,…,P,-}, where – (minus) designates empty grid cells. The data sets are separated by single empty lines and terminate with an end of file. Output The program prints the solution of the input encoded grids in the same format and order as used for input. Sample Input --A----C-----O-I -J--A-B-P-CGF-H- --D--F-I-E----P- -G-EL-H----M-J-- ----E----C--G--- -I--K-GA-B---E-J D-GP--J-F----A-- -E---C-B--DP--O- E--F-M--D--L-K-A -C--------O-I-L- H-P-C--F-A--B--- ---G-OD---J----H K---J----H-A-P-L --B--P--E--K--A- -H--B--K--FI-C-- --F---C--D--H-N- Sample Output FPAHMJECNLBDKOGI OJMIANBDPKCGFLHE LNDKGFOIJEAHMBPC BGCELKHPOFIMAJDN MFHBELPOACKJGNID CILNKDGAHBMOPEFJ DOGPIHJMFNLECAKB JEKAFCNBGIDPLHOM EBOFPMIJDGHLNKCA NCJDHBAEKMOFIGLP HMPLCGKFIAENBDJO AKIGNODLBPJCEFMH KDEMJIFNCHGAOPBL GLBCDPMHEONKJIAF PHNOBALKMJFIDCEG IAFJOECGLDPBHMNK Source Southeastern Europe 2006

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

using namespace std;

const int N=16;
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)/16]=(ans[i]-1)%16+'A';
			//printf("%s\n",sudoku);
			for(int i=0,sz=N*N;i<sz;i++)
            {
                putchar(sudoku[i]);
                if((i+1)%16==0) putchar(10);
            }
			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/4)*4+(j/4))*N+k;
}

int main()
{
	while(scanf("%s",sudoku)!=EOF)
	{
		for(int i=1;i<16;i++)
            scanf("%s",sudoku+i*16);
		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+'A'-1)
					{
						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);
		putchar(10);
	}
	return 0;
}

POJ 3076 Sudoku DLX精确覆盖

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原文地址：http://blog.csdn.net/ck_boss/article/details/40056621