2015南阳理工CCPC Sudoku

Sudoku

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×2 pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!

Input

The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4characters. Each character represents the number in the corresponding cell (one of 1, 2, 3, 4). * represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.

Sample input and output

Sample InputSample Output

3****234141233214*243*312*421*134*41***3*2*414*2*

Case #1:1432234141233214Case #2:1243431234212134Case #3:3412123423414123

Source

The 2015 China Collegiate Programming Contest

题解: 这题爆搜即可过, 我们可以设三个标记数组, vis1[i][j] 表示第i行是否出现过j, vis2[i][j]表示第i列是否出现过j, vis3[i][j] 表示四个小区域中是否出现过j

然后用dfs搜索即可.

#include <cstdio>#include <cstring>const int N = 6;char mp[N][N];bool vis1[N][N], vis2[N][N], vis3[N][N];bool found;void search(int cur) {if (cur == 16) {found = true;return ;}int r = cur / 4, c = cur % 4;int idx = mp[r][c] - '0';if (mp[r][c] == '*') {for (int i = 1; i <= 4; ++i) {if (!vis1[r][i] && !vis2[c][i] && !vis3[r / 2 * 2 + c / 2][i]) {vis1[r][i] = vis2[c][i] = vis3[r / 2 * 2 + c / 2][i] = true;mp[r][c] = '0' + i;search(cur + 1);if (found) break;mp[r][c] = '*';vis1[r][i] = vis2[c][i] = vis3[r / 2 * 2 + c / 2][i] = false;}}}else search(cur + 1);}void init() {memset(vis1, false, sizeof(vis1));memset(vis2, false, sizeof(vis2));memset(vis3, false, sizeof(vis3));found = false;}int main() {int T;scanf("%d", &T);while (T--) {init();for (int i = 0; i < 4; ++i) {scanf("%s", mp[i]);for (int j = 0; j < 4; ++j)if (mp[i][j] != '*') {int idx = mp[i][j] - '0';vis1[i][idx] = vis2[j][idx] = vis3[i / 2 * 2 + j / 2][idx] = true;}}search(0);static int cnt = 0;printf("Case #%d:\n", ++cnt);for (int i = 0; i < 4; ++i)printf("%s\n", mp[i]);}return 0;}