题目链接:UVA 11210 中国麻将
题意:一般麻将的规则,给出13张手牌,按指定顺序输出可以“听”的牌,没有听牌输出“Not ready”;
题解:这题和UVA 11464偶数矩阵的思路很像,在数量级小且无法判断的情况下进行必要的枚举以达到求解条件。
AC code:
//lrl's submission //adrui 's source code for UVA 11210 //Memory: 0 KB Time : 10 MS //Language : C++ 5.3.0 Result : Accepted #include<cstdio> #include<cstring> #define M(a, b) memset(a, b, sizeof(a)) #define debug 0 const int maxn = 100000 + 5; int num[34],mj[15]; char s[10]; char *majiang[] = { "1T","2T", "3T", "4T", "5T", "6T", "7T", "8T", "9T", "1S","2S","3S", "4S", "5S", "6S", "7S", "8S", "9S", "1W","2W" ,"3W", "4W", "5W", "6W", "7W", "8W", "9W","DONG", "NAN", "XI", "BEI", "ZHONG", "FA", "BAI" }; //34种手牌 int find(char *s) { for (int i = 0; i < 34; i++) { if (strcmp(majiang[i], s) == 0) return i; //查找下标 } return -1; } bool test(int depth) { for (int i = 0; i < 34; i++) //对子 if(num[i] >= 3){ if (depth == 3)return true; //去掉两张将牌,只要有四组顺牌就可 num[i] -= 3; if (test(depth + 1)) return true; //递归,有点dfs的意思 num[i] += 3; //状态返回 } for(int i = 0; i <= 24;i++) if (i % 9 <= 6 && num[i] >= 1 && num[i + 1] >= 1 && num[i + 2] >= 1)//顺子 { if (depth == 3)return true; num[i]--; num[i + 1]--; num[i + 2]--; if (test(depth + 1)) return true; num[i]++; num[i + 1]++; num[i + 2]++; } return false; } bool check(){ for (int i = 0; i < 34; i++) { if (num[i] >= 2) { //枚举将牌的情况 num[i] -= 2; if (test(0)) return true; num[i] += 2; //返回状态 } } return false; } int main() { #if debug freopen("in.txt", "r", stdin); #endif//debug int kase = 1; while (~scanf("%s", s)) { if (s[0] == '0') break; mj[0] = find(s); for (int i = 1; i < 13; i++) { scanf("%s", s); mj[i] = find(s); } /*for (int j = 0; j < 13; j++) printf("%3d", mj[j]); puts("");*/ printf("Case %d:", kase++); int flag = 0; for (int i = 0; i < 34; i++) { //枚举34种牌 M(num, 0); //不能少 for (int j = 0; j < 13; j++) num[mj[j]]++; if (num[i] < 4) { num[i]++; //14张手牌。 if (check()) { flag = 1; printf(" %s", majiang[i]); } num[i]--; } } if (!flag) printf(" Not ready"); puts(""); } return 0; }
HDU 4431
AC code:(这题有十三幺。。没玩过和7pair的和牌法)
//Problem: 4431 (Mahjong)Judge Status : Accepted //RunId : 18164713 Language : G++ Author : adruill //Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta #include<cstdio> #include<cstring> #define M(a, b) memset(a, b, sizeof(a)) #define debug 0 const int maxn = 100000 + 5; int num[34],mj[15],ans[34]; char s[10]; char *majiang[] = { "1m","2m", "3m", "4m", "5m", "6m", "7m", "8m", "9m","1s","2s" ,"3s", "4s", "5s", "6s", "7s", "8s", "9s", "1p","2p","3p", "4p", "5p", "6p", "7p", "8p", "9p", "1c", "2c", "3c", "4c", "5c", "6c", "7c" }; int find(char *s) { for (int i = 0; i < 34; i++) { if (strcmp(majiang[i], s) == 0) return i; } return -1; } bool is7() { for (int i = 0; i < 34; i++) { if (num[i] && num[i] != 2) return false; } return true; } bool is131() { if (num[0] && num[8] && num[9] && num[17] && num[18] && num[26] && num[27] && num[28] && num[29] && num[30] && num[31] && num[32] && num[33]) { if (num[0] + num[8] + num[9] + num[17] + num[18] + num[26] + num[27] + num[28] + num[29] + num[30] + num[31] +num[32] + num[33] == 14) return true; } return false; } bool test() { int ret = 0; int tmp[35]; for (int i = 0; i<34; i++)tmp[i] = num[i]; for (int i = 0; i <= 18; i += 9) for (int j = 0; j < 9; j++) { if (tmp[i + j] >= 3) { tmp[i + j] -= 3; ret++; } while (j + 2<9 && tmp[i + j] && tmp[i + j + 1] && tmp[i + j + 2]) { tmp[i + j]--; tmp[i + j + 1]--; tmp[i + j + 2]--; ret++; } } for (int j = 0; j < 7; j++) { if (tmp[27 + j] >= 3) { tmp[27 + j] -= 3; ret++; } } if (ret == 4)return true; return false; } bool check(){ for (int i = 0; i < 34; i++) { if (is7() || is131()) return true; if (num[i] >= 2) { num[i] -= 2; if (test()) return true; num[i] += 2; } } return false; } int main() { #if debug freopen("in.txt", "r", stdin); #endif//debug int t; scanf("%d", &t); while (t--) { M(ans, 0); for (int i = 0; i < 13; i++) { scanf("%s", s); mj[i] = find(s); } for (int i = 0; i < 34; i++) { M(num, 0); for (int j = 0; j < 13; j++) num[mj[j]]++; if (num[i] < 4) { num[i]++; if (check()) { ans[0]++; ans[ans[0]] = i; } num[i]--; } } if (!ans[0]) printf("Nooten"); else { printf("%d", ans[0]); for (int i = 1; i <= ans[0]; i++) printf(" %s", majiang[ans[i]]); } puts(""); } return 0; }