UVa989 - Su Doku(数独游戏)

In many newspapers we may nd some puzzles to solve, one of those is Su Doku. Given a grid 9  9
with some of entries lled, the objective is to ll in the grid so that every row, every column, and every
3  3 box contains the digits 1 through 9.
source: http://www.sudoku.com
Input
Input contains several test cases separated by a blank line. Each of them contains an integer n such
that 1  n  3 and a grid n
2  n
2
with some of the entries lled with digits from 1 to n
2
(an entrie
not lled will have 0). In this case, the objective is to ll in the grid so that every row, every column,
and every n  n box contains the digits 1 through n
2
.
Output
A solution for the problem. If exists more than one, you should give the lower one assuming a lexico-
graphic order. If there is no solution, you should print `NO SOLUTION'. For lexicographic comparison
you should consider lines in rst place. Print a blank line between test cases.
Sample Input
3
0 6 0 1 0 4 0 5 0
0 0 8 3 0 5 6 0 0
2 0 0 0 0 0 0 0 1
8 0 0 4 0 7 0 0 6
0 0 6 0 0 0 3 0 0
7 0 0 9 0 1 0 0 4
5 0 0 0 0 0 0 0 2
0 0 7 2 0 6 9 0 0
0 4 0 5 0 8 0 7 0
Sample Output
9 6 3 1 7 4 2 5 8
1 7 8 3 2 5 6 4 9
2 5 4 6 8 9 7 3 1
8 2 1 4 3 7 5 9 6
4 9 6 8 5 2 3 1 7
7 3 5 9 6 1 8 2 4
5 8 9 7 1 3 4 6 2
3 1 7 2 4 6 9 8 5
6 4 2 5 9 8 1 7 3

#include <cstdio>#include <cstring>#include <vector>#include <set>using namespace std;const int N = 9;int vis[N][N];int n;int g[N][N];set<int> s;int cas = 0;void init(){    for (int i = 1; i <= n * n; i++) {        s.insert(i);    }}bool input(){    if (scanf("%d", &n) != 1) return false;    //printf("n=%d\n", n);    int m = n * n;    s.clear();    init();    memset(vis, 0x00, sizeof(vis));    for (int i = 0; i < m; i++) {        for (int j = 0; j < m; j++) {            scanf("%d", &g[i][j]);            if (g[i][j]) {                vis[i][j] = g[i][j];            }        }    }#if 0    for (int i = 0; i < m; i++) {        for (int j = 0; j < m; j++) {            printf(" %d", g[i][j]);        }        printf("\n");    }#endif    return true;}bool dfs(int row, int col){    if (row == n * n) return true;    if (g[row][col]) {        col++;        if (col >= n * n) {            row++, col = 0;        }        return dfs(row, col);    } else {        set<int> tmp = s;        for (int i = 0; i < n * n; i++) {            if (i != col && vis[row][i]) {                tmp.erase(vis[row][i]);            }        }        for (int i = 0; i < n * n; i++) {            if (i != row && vis[i][col]) {                tmp.erase(vis[i][col]);            }        }        int x1 = row / n, y1 = col / n;        for (int i = x1 * n; i < (x1 + 1) * n; i++) {            for (int j = y1 * n; j < (y1 + 1) * n; j++) {                if (i != row && j != col && vis[i][j]) tmp.erase(vis[i][j]);            }        }        if (tmp.empty()) return false;#if 0        cout << "row:" << row << " col:" << col << " tmp:";        copy(tmp.begin(), tmp.end(), ostream_iterator<int>(cout, " "));        cout << endl;#endif        for (set<int>::iterator it = tmp.begin(); it != tmp.end(); it++) {            g[row][col] = *it;            vis[row][col] = *it;            if (dfs(row, col)) return true;            g[row][col] = 0;            vis[row][col] = 0;        }        return false;    }}void solve(){    if (cas++) printf("\n");    if (dfs(0, 0)) {        for (int i = 0; i < n * n; i++) {            for (int j = 0; j < n * n; j++) {                if (j) printf(" ");                printf("%d", g[i][j]);            }            printf("\n");        }    } else {        printf("NO SOLUTION\n");    }}int main(){#ifndef ONLINE_JUDGE    freopen("e:\\uva_in.txt", "r", stdin);#endif    while (input()) {        solve();    }    return 0;}