acm algorithm practice Jan 11 DFS

poj 2676

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. 

-----------------------------------------------

a classical DFS problem search from the rear of the table.  

代码

#include < iostream >

using namespace std;

char num[ 9 ][ 9 ];

void Init()
{
int i, k;
for (i = 0 ; i < 9 ; i ++ )
{
for (k = 0 ; k < 9 ; k ++ )
{
cin >> num[i][k];
}
}
}

inline bool RowValid( int row) // 检测行冲突
{
int k;
bool has[ 60 ] = { false };
for (k = 0 ; k < 9 ; k ++ )
{
if (num[row][k] != ' 0 ' )
{
if (has[num[row][k]]) // in the array
return false ;
else // not in the array in the array
has[num[row][k]] = true ;
}
}
return true ;
}

inline bool ColValid( int col) // 检测列合法性
{
int i;
bool has[ 60 ] = { false };
for (i = 0 ; i < 9 ; i ++ )
{
if (num[i][col] != ' 0 ' )
{
if (has[num[i][col]])
return false ;
else
has[num[i][col]] = true ;
}
}
return true ;
}

inline bool BlockValid( int brow, int bcol) // 3×3的合法行
{
int i, k;
bool has[ 60 ] = { false };
int startrow = brow * 3 ; // find the start row of the block
int startcol = bcol * 3 ; // start col of the block
for (i = startrow; i < startrow + 3 ; i ++ )
{
for (k = startcol; k < startcol + 3 ; k ++ )
{
if (num[i][k] != ' 0 ' ) // caveat 1 don't miss
{
if (has[num[i][k]])
return false ;
else
has[num[i][k]] = true ;
}
}
}
return true ;
}

inline bool Valid( int row, int col)
{
if ( ! RowValid(row)) // judge the row
return false ;
if ( ! ColValid(col)) // judge the col
return false ;
if ( ! BlockValid(row / 3 , col / 3 )) // judge the block
return false ;
return true ;
}

inline bool DFS( int pos) // 深搜，这样的单参数比较2个参数（行，列）方便。
{
if ( - 1 == pos)
return true ; // DFS is finished

int row = pos / 9 ;
int col = pos % 9 ;

if (num[row][col] != ' 0 ' ) // the block is already filled, move to next
return DFS(pos - 1 );

char i;
for (i = ' 1 ' ; i <= ' 9 ' ; i ++ )
{
num[row][col] = i; // probe the number from 1 to 9
if (Valid(row, col)) // if it pass the judge move to the next
{
if (DFS(pos - 1 )) // all the sub-DFS have returned true
return true ;
}
num[row][col] = ' 0 ' ; // traceback
}
return false ; // only when all the judges are false, actually not necessary
}

void Print()
{
int i, k;
for (i = 0 ; i < 9 ; i ++ )
{
for (k = 0 ; k < 9 ; k ++ )
{
printf( " %c " , num[i][k]);
}
printf( " \n " );
}
}

int main()
{
int cases;
scanf( " %d " , & cases);
while (cases -- )
{
Init();
DFS( 80 );
Print();
}
return 0 ;
}

 

 tips:

1,  probe and traceback method 

 DFS (int n){   //which means  from 1 to n blocks are done

        // ........

       // judge if the nth block is ok

       DFS(n - 1)   //which means  from 1 to n-1 blocks are down 

       //now we can return true since nth depends on n-1th 

       //.......

       //if the nth is not okay, trace back

}

2, u should search from the rear rather than from the head, or it will cause TLE( because of the special judge. )

3, in DFS use single parameter rather than x, y since the dfs is moving downward in order. 

4, dont convert char to int, it will cost extra time and lead to TLE. 

6, use int , dfs, print 3 functions to make the code clean. 

 

obviously, it's  NP problem 

转载于:https://www.cnblogs.com/ggppwx/archive/2011/01/12/1933742.html