题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1426
思路分析:该问题为数独问题,明显解是唯一的,所有采用dfs搜索效果更好;
在搜索时,可以通过3个数组来判断对于某个特定的数是否能够满足要求,即在每一行、每一列和每一个3X3的方块中只有唯一的1~9之间的数;
vis_r数组:如果vis_r[i][j] == 1表示在第i行中数字j已经存在,否则表示不存在;
vis_c数组:如果vis_c[i][j] == 1表示在第j列数字j已经存在,否则表示不存在;
对于3X3方块,按照从左到右,从上到下的顺序编号为0~9;
如果vis_k[i][j] == 1则表示第i号方块中数字j已经存在,否则表示不存在;
代码如下:
#include <cstdio>
#include <iostream>
using namespace std; const int MAX_N = 85;
const int MAX_M = 10;
int map[MAX_M][MAX_M];
int vis_r[MAX_M][MAX_M];
int vis_c[MAX_M][MAX_M];
int vis_k[MAX_M][MAX_M];
struct Node { int x, y; };
Node arr[MAX_N]; inline int ChunkNumber(int i, int j) { return i / 3 * 3 + j / 3; }
void Dfs(int deep, int k, int &find_ans)
{
if (deep == k)
{
find_ans = 1;
return;
}
else
{
int x = arr[deep].x;
int y = arr[deep].y;
int num_chunk = ChunkNumber(x, y); for (int i = 1; i <= 9; ++i)
{
if (!vis_r[x][i] && !vis_c[y][i] && !vis_k[num_chunk][i])
{ map[x][y] = i;
vis_r[x][i] = 1;
vis_c[y][i] = 1;
vis_k[num_chunk][i] = 1;
Dfs(deep + 1, k, find_ans);
if (find_ans) return;
vis_r[x][i] = 0;
vis_c[y][i] = 0;
vis_k[num_chunk][i] = 0;
map[x][y] = 0;
}
}
}
} int main()
{
char temp[MAX_M][MAX_M];
int k = 0, find_ans = 0;
int flag = 0; while (scanf("%s", &temp[0][0]) != EOF)
{
k = find_ans = 0;
for (int j = 1; j < 9; ++j)
scanf("%s", &temp[0][j]);
for (int i = 1; i < 9; ++i)
for (int j = 0; j < 9; ++j)
scanf("%s", &temp[i][j]); memset(vis_r, 0, sizeof(vis_r));
memset(vis_c, 0, sizeof(vis_c));
memset(vis_k, 0, sizeof(vis_k));
for (int i = 0; i < 9; ++i)
{
for (int j = 0; j < 9; ++j)
{
if (temp[i][j] == '?')
{
map[i][j] = 0;
arr[k].x = i;
arr[k].y = j;
k++;
}
else
{
int num = 0; num = map[i][j] = temp[i][j] - '0';
vis_r[i][num] = 1;
vis_c[j][num] = 1;
vis_k[ChunkNumber(i, j)][num] = 1;
}
}
}
Dfs(0, k, find_ans); if (flag)
printf("\n");
else
flag = 1;
for (int i = 0; i < 9; ++i)
for (int j = 0; j < 9; ++j)
{
if (j != 8)
printf("%c ", map[i][j] + '0');
else
printf("%c\n", map[i][j] + '0');
}
}
return 0;
}