POJ 2676 Sudoku 解题报告（Dancing Link）

Sudoku Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13019   Accepted: 6452   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

   

    解题报告： 和POJ 3074是一样的，稍微改下代码即可。代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int M=1024*110;
const int N=1024;

int l[M], r[M], d[M], u[M], col[M], row[M], h[N], control[N];
int dcnt = 0;
char str[111];

inline void addnode(int &x)
{
	++x;
	r[x]=l[x]=u[x]=d[x]=x;
}

inline void insert_row(int rowx, int x)
{
	r[l[rowx]]=x;
	l[x]=l[rowx];
	r[x]=rowx;
	l[rowx]=x;
}

inline void insert_col(int colx, int x)
{
	d[u[colx]]=x;
	u[x]=u[colx];
	d[x]=colx;
	u[colx]=x;
}

void dlx_init(int cols)
{
	memset(h, -1, sizeof(h));
	memset(control, 0, sizeof(control));
	dcnt=-1;
	addnode(dcnt);

	for(int i=1;i<=cols;++i)
	{
		addnode(dcnt);
		insert_row(0, dcnt);
	}
}

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];
			control[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])
		{
			u[d[j]]=j;
			d[u[j]]=j;
			control[col[j]]++;
		}
		l[r[c]]=c;
		r[l[c]]=c;
}

bool DLX(int deep)
{
	if(r[0]==0)
	{
		for(int i=0;i<9;i++)
		{
			for(int j=0;j<9;j++)
				printf("%c", str[i*9+j]);
			puts("");
		}
		return true;
	}

	int min=M, tempc;
	for(int i=r[0];i!=0;i=r[i]) if(control[i]<min)
	{
		min=control[i];
		tempc=i;
	}
	remove(tempc);
	for(int i=d[tempc];i!=tempc;i=d[i])
	{
		str[row[i]/9]=row[i]%9+'1';
		for(int j=r[i];j!=i;j=r[j]) remove(col[j]);
		if(DLX(deep+1)) return true;
		for(int j=l[i];j!=i;j=l[j]) resume(col[j]);
	}
	resume(tempc);
	return false;
}

inline void insert_node(int x, int y)
{
	control[y]++;
	addnode(dcnt);
	row[dcnt]=x;
	col[dcnt]=y;
	insert_col(y, dcnt);
	if(h[x]==-1)
		h[x]=dcnt;
	else
		insert_row(h[x], dcnt);
}

int main()
{
#ifdef ACM
	freopen("in.txt", "r", stdin);
#endif

	int T;
	scanf("%d", &T);

	while(T--)
	{
		for(int i=0;i<9;i++) scanf("%s", str+9*i);

		dlx_init(4*9*9);
		for(int i=1;i<=9;i++) for(int j=1;j<=9;j++)
		{
			for(int k=1;k<=9;k++) if(str[(i-1)*9+(j-1)]=='0' || str[(i-1)*9+(j-1)]=='0'+k)
			{
				int rr=(i-1)*9*9+(j-1)*9+(k-1);
				insert_node(rr, 81*0+(i-1)*9+k);
				insert_node(rr, 81*1+(j-1)*9+k);
				insert_node(rr, 81*2+((i-1)/3+(j-1)/3*3)*9+k);
				insert_node(rr, 81*3+(i-1)*9+j);
			}
		}

		DLX(0);
	}
}