poj2676 Sudoku

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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

Output

Sample Input

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

这道题可以用深搜，题意是补完数独中的空格数，0代表空格。用了三个剪枝，同行，同列还有同一个大格子的数字不能重复。

#include<stdio.h>
#include<string.h>
char s[12][12];
int row[12][12],col[12][12],dage[12][12],map[12][12],flag;
void dfs(int x,int y)
{
	int i,j,k;
	if(x==10){
		flag=1;return;
	}
	if(map[x][y]!=0){
		if(y==9){
			dfs(x+1,1);
		}
		else dfs(x,y+1);
		if(flag)return;
	}
	else if(map[x][y]==0){
	    k=(x-1)/3*3+(y-1)/3+1;
		for(i=1;i<=9;i++){
			if(row[x][i]==0 && col[y][i]==0 && dage[k][i]==0){
				map[x][y]=i;
				row[x][i]=1;col[y][i]=1;dage[k][i]=1;
				if(y==9){
					dfs(x+1,1);
				}
				else dfs(x,y+1);
			    if(flag==1)return;
			    map[x][y]=0;row[x][i]=0;col[y][i]=0;dage[k][i]=0;
			}
		}
	}
}


int main()
{
	int n,m,i,j,T,k,flag1;
	scanf("%d",&T);
	while(T--)
	{
		memset(row,0,sizeof(row));memset(col,0,sizeof(col));memset(dage,0,sizeof(dage));memset(map,0,sizeof(map));
		flag1=1;
		for(i=1;i<=9;i++){
			scanf("%s",s[i]+1);
			for(j=1;j<=9;j++){
				map[i][j]=s[i][j]-‘0‘;
				k=(i-1)/3*3+(j-1)/3+1;
				if(map[i][j]!=0){
					if(row[i][map[i][j]]!=0 || col[j][map[i][j]]!=0 || dage[k][map[i][j]]!=0){
						flag1=0;break;
					}
					row[i][map[i][j]]=1;
					col[j][map[i][j]]=1;
					dage[k][map[i][j]]=1;
				}
			}
			if(flag1==0)break;
		}
		if(flag1==0){
			for(i=1;i<=9;i++){
				printf("%s\n",s[i]+1);
			}
		}
		flag=0;
		dfs(1,1);
		if(flag==0){
			for(i=1;i<=9;i++){
				printf("%s\n",s[i]+1);
			}
		}
		else{
			for(i=1;i<=9;i++){
				for(j=1;j<=9;j++){
					printf("%d",map[i][j]);
				}
				printf("\n");
			}
		}
	}
	return 0;
}

poj2676 Sudoku

标签：搜索

原文地址：http://blog.csdn.net/kirito_acmer/article/details/45533097