| Time limit : 1 sec | Memory limit : 64 M |
Tom is a master in several mathematical-theoretical disciplines. He recently founded a research-lab at ouruniversity and teaches newcomers like Jim. In the first lesson he explained the game of Tudoku to Jim.Tudoku is a straight-forward variant of Sudoku, because it consists of a board where almost all the numbersare already in place. Such a board is left over when Tom stops solving an ordinary Sudoku because of beingtoo lazy to fill out the last few straight-forward cells. Now, you should help Jim solve all Tudokus Tom leftfor him.
Problem
Sudoku is played on a 9 * 9 board that is divided into nine different 3 * 3 blocks. Initially, it containsonly a few numbers and the goal is to fill the remaining cells so that each row, column, and 3 * 3 blockcontains every number from 1 to 9. This can be quite hard but remember that Tom already filled most cells.A resulting Tudoku board can be solved using the following rule repeatedly: if some row, column or 3 * 3block contains exactly eight numbers, fill in the remaining one.
In the following example, three cells are still missing. The upper left one cannot be determined directlybecause neither in its row, column, or block, there are eight numbers present. The missing number forthe right cell can be determined using the above rule, however, because its column contains exactly eightnumbers. Similarly, the number for the lower-most free cell can be determined by examining its row.Finally, the last free cell can be filled by either looking at its row, column or block.
Input
The first line contains the number of scenarios. For each scenario the input contains nine lines of nine digitseach. Zeros indicate the cells that have not been filled by Tom and need to be filled by you. Each scenariois terminated by an empty line.
Output
The output for every scenario begins with a line containing "Scenario #i:", where i is the number ofthe scenario starting at 1. Then, print the solved Tudoku board in the same format that was used for theinput, except that zeroes are replaced with the correct digits. Terminate the output for the scenario with ablank line.
Sample Input
2 000000000 817965430 652743190 175439820 308102950 294856370 581697240 903504610 746321580 781654392 962837154 543219786 439182675 158976423 627543918 316728549 895461237 274395861
Sample Output
Scenario #1: 439218765 817965432 652743198 175439826 368172954 294856371 581697243 923584617 746321589 Scenario #2: 781654392 962837154 543219786 439182675 158976423 627543918 316728549 895461237 274395861
Solution:
#include <stdio.h>
//分別計算行、列、塊中0的個數（即需要填寫位置的個數），並返回總的0的個數
int count0(int a[][9],int count[][9])
{
int m = 0,count1 = 0;
for(int i=0;i<3;i++)
{
for(int j=0;j<9;j++)
{
count[i][j] = 0;
}
}
for(int i=0;i<9;i++)
{
for(int j=0;j<9;j++)
{
m = i/3*3+j/3;
if(a[i][j] == 0)
{
count[0][i]++;
count[1][j]++;
count[2][m]++;
count1++;
}
}
}
return count1;
}
void clear(int a[])
{
for(int i=0;i<9;i++)
{
a[i] = 0;
}
}
void todoku(int a[][9])
{
int count[3][9] = {0};
int pos1,pos2,m,n,flag=0;
int number[9] = {0};
while(count0(a,count))
{
for(int i=0;i<9;i++)
{
//處理0個數爲1的行
if(count[0][i] == 1)
{
for(int j=0;j<9;j++)
{
if(a[i][j] == 0)
{
pos1 = j;
flag=1;
}
else
number[a[i][j]-1] = 1;
}
if(flag)
{
for(int j=0;j<9;j++)
{
if(number[j] == 0)
a[i][pos1] = j+1;
}
clear(number);
flag = 0;
}
}//end if
//處理0個數爲1的列
if(count[1][i] == 1)
{
for(int j=0;j<9;j++)
{
if(a[j][i] == 0)
{
pos1 = j;
flag = 1;
}
else
{
number[a[j][i]-1] = 1;
}
}
if(flag)
{
for(int j=0;j<9;j++)
{
if(number[j] == 0)
a[pos1][i] = j+1;
}
clear(number);
flag = 0;
}
}//end if
//處理0個數爲1的塊
if(count[2][i] == 1)
{
m=i/3*3;
n=i%3*3;
for(int j=0;j<3;j++)
{
for(int k=0;k<3;k++)
{
if(a[m+j][n+k] == 0)
{
pos1=m+j;
pos2=n+k;
flag = 1;
}
else
number[a[m+j][n+k]-1] = 1;
}
}
if(flag)
{
for(int j=0;j<9;j++)
{
if(number[j] == 0)
{
a[pos1][pos2] = j+1;
}
}
clear(number);
flag = 0;
}
}//end if
}//end for
}//end while
}
int main()
{
int num,a[9][9]={0};
scanf("%d",&num);
for(int i=1;i<=num;i++)
{
for(int m=0;m<9;m++)
{
for(int n=0;n<9;n++)
{
scanf("%1d",&a[m][n]);
}
}
todoku(a);
printf("Scenario #%d:\n",i);
for(int m=0;m<9;m++)
{
for(int n=0;n<9;n++)
{
printf("%d",a[m][n]);
}
printf("\n");
}
printf("\n");
}
}
//只是能實現，算法不是最好，大家多提意見