11210-Chinese Mahjong (dfs)

Question: 11210-Chinese Mahjong

I will give 13 Mahjong members and ask which one can be used? List all the situations in the order required by the questions. Hu's condition requires one and only one pair, and then there are either three identical or three consecutive (the premise is that the suffix is the same and only T, S, W is in consideration)

Solution: enumerate the situations to be retrieved one by one, and then dfs. You can check if you want to add them. If you want to find them, you can check the three cases. However, note that the suffixes must be the same for the three consecutive periods. Here, I convert Mahjong into numbers and count the number of times this Mahjong appears. Therefore, some boundary judgments are required.

Code:

#include 
 
  #include 
  
   const int N = 35;const char s[7][7] = {"DONG", "NAN", "XI", "BEI", "ZHONG", "FA", "BAI"};int tile[N];char str[N];int handle () {if (str[0] >= '1' && str[0] <= '9') {int add;switch (str[1]) {case 'T': add = 0;break;case 'S': add = 9;break;case 'W': add = 18;break;}return add + str[0] - '0';} else {int i;for (i = 0; i < 7; i++)if (strcmp(str, s[i]) == 0)break;return 28 + i;}}bool dfs (int k, int flag) {if (k >= 14) {if (flag)return true;return false;}for (int i = 1; i < N; i++) {if (tile[i] >= 2 && !flag) {tile[i] -= 2;if (dfs (k + 2, 1)) {tile[i] += 2;return true;} tile[i] += 2;}if (tile[i] >= 3) {tile[i] -= 3;if (dfs (k + 3, flag)) {tile[i] += 3;return true;}tile[i] += 3;}if ( (i > 7&& i < 10) || (i > 16 && i < 19) || i > 25)continue;if (tile[i] >= 1 && tile[i + 1] >= 1 && tile[i + 2] >= 1) {tile[i]--;tile[i + 1]--;tile[i + 2]--;if (dfs (k + 3, flag)) {tile[i]++;tile[i + 1]++;tile[i + 2]++;return true;}tile[i]++;tile[i + 1]++;tile[i + 2]++;}}return false;}int main () {int ans[N], k, cas = 0;while (scanf ("%s", str)) {memset (tile, 0, sizeof(tile));k = 0;if (str[0] == '0')break;tile[handle()]++;for (int i = 0; i < 12; i++) {scanf ("%s", str);tile[handle()]++;}for (int i = 1; i < N; i++) {if (tile[i] >= 4)continue;tile[i]++;if (dfs (0, 0))ans[k++] = i;tile[i]--;}printf ("Case %d: ", ++cas);if (!k) {printf ("Not ready\n");continue;}for (int i = 0; i < k; i++) {if (i < k - 1) {if (ans[i] >= 1 && ans[i] <= 9)printf ("%dT ", ans[i]);else if (ans[i] >= 10 && ans[i] <= 18)printf ("%dS ", ans[i]- 9);else if (ans[i] >= 19 && ans[i] <= 27)printf ("%dW ", ans[i] - 18);elseprintf ("%s ", s[ans[i] - 28]);} else {if (ans[i] >= 1 && ans[i] <= 9)              printf ("%dT\n", ans[i]);else if (ans[i] >= 10 && ans[i] <= 18)          printf ("%dS\n", ans[i] - 9);else if (ans[i] >= 19 && ans[i] <= 27)printf ("%dW\n", ans[i] - 18);   elseprintf ("%s\n", s[ans[i] - 28]);}}}}