UVa 989

题目：数独。填充大小为1x1，4x4，9x9数独。

分析：搜索。直接利用dfs回溯求解。存储每行，每列，每个子块的数字使用情况，优化搜索。

说明：对于解有顺序要求，DL超时了 - -。

#include <cstring>#include <cstdio>int data[9][9];int b_id[9][9];int visit_r[10][10];int visit_c[10][10];int visit_b[10][10];int position[81];int dfs(int k, int size, int n){if (k == size) {return 1;}int p = position[k];int r = p/(n*n), c = p%(n*n), b = r/n*n + c/n;for (int v = 1; v <= n*n; ++ v) {if (!visit_r[r][v] && !visit_c[c][v] && !visit_b[b][v]) {visit_r[r][v] = 1;visit_c[c][v] = 1;visit_b[b][v] = 1;data[r][c] = v;if (dfs(k+1, size, n)) {return 1;}visit_r[r][v] = 0;visit_c[c][v] = 0;visit_b[b][v] = 0;}}return 0;} int main(){int n, t = 0;while (~scanf("%d",&n)) {if (t ++) {printf("\n");}for (int i = 0; i < n*n; ++ i) {for (int v = 1; v <= n*n; ++ v) {visit_r[i][v] = 0;visit_c[i][v] = 0;visit_b[i][v] = 0;}}int size = 0;for (int i = 0; i < n*n; ++ i) {for (int j = 0; j < n*n; ++ j) {scanf("%d",&data[i][j]);b_id[i][j] = i/n*n + j/n;if (data[i][j] == 0) {position[size ++] = i*n*n + j;}else {visit_r[i][data[i][j]] = 1;visit_c[j][data[i][j]] = 1;visit_b[b_id[i][j]][data[i][j]] = 1;}}}if (dfs(0, size, n)) {for (int i = 0; i < n*n; ++ i) {for (int j = 0; j < n*n; ++ j) {if (j) {printf(" ");}printf("%d",data[i][j]);}printf("\n");}}else {printf("NO SOLUTION\n");}}return 0;}