UVA 11210 暴力枚举 + 递归(hdu 4431)

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 KBTime : 10 MS//Language : C++ 5.3.0Result : Accepted#include<cstdio>#include<cstring>#define M(a, b) memset(a, b, sizeof(a))#define debug 0const 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 debugfreopen("in.txt", "r", stdin);#endif//debugint 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 0const 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 debugfreopen("in.txt", "r", stdin);#endif//debugint 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;}