Chinese Mahjong

OJ题号：UVa11210

思路：

首先字符串处理读入手牌，str数组将手牌和数字对应，接下来搜索，先搜对子，如果搜过对子就不搜了。由于对子有且只有一个，可以在搜到以后直接跳出。同时注意一副麻将中每种牌只有4个，所以不能作为将。

 #include<cstdio>
 #include<string>
 #include<vector>
 #include<sstream>
 #include<iostream>
 using namespace std;
 vector<int> mahjong;
 const string str[]={"","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"};
 void input(string line,vector<int> &ans) {
     ans.clear();
     for(int i=;i<;i++) ans.push_back();
     stringstream ss(line);
     string temp;
     for(int i=;i<;i++) {
         ss>>temp;
         for(int j=;j<=;j++) {
             if(temp==str[j]) {
                 ans[j]++;
                 break;
             }
         }
     }
 }
 bool duizi=,flag=;
 int ting;
 bool hu() {
     for(int i=;i<=;i++) {
         if(mahjong[i]) {
             goto Try;
         }
     }
     cout<<" "+str[ting];
     flag=;
     return ;
     Try:
     if(!duizi) {
         for(int i=;i<=;i++) {
             if(mahjong[i]>=) {
                 mahjong[i]-=;
                 duizi=;
                 bool f=hu();
                 duizi=;
                 mahjong[i]+=;
                 if(f) return ;
             }
         }
         return ;
     }
     for(int i=;i<=;i++) {
         if(mahjong[i]>=) {
             mahjong[i]-=;
             bool f=hu();
             mahjong[i]+=;
             if(f) return ;
         }
     }
     for(int i=;i<=;i++) {
         if(i==||i==||i==||i==||i==||i==) continue;
         if(mahjong[i]&&mahjong[i+]&&mahjong[i+]) {
             mahjong[i]--;
             mahjong[i+]--;
             mahjong[i+]--;
             bool f=hu();
             mahjong[i]++;
             mahjong[i+]++;
             mahjong[i+]++;
             if(f) return ;
         }
     }
     return ;
 }
 int main() {
     int turn=;
     while() {
         string s;
         getline(cin,s);
         if(s=="") break;
         turn++;
         printf("Case %d:",turn);
         mahjong.clear();
         input(s,mahjong);
         flag=;
         for(ting=;ting<=;ting++) {
             if(mahjong[ting]==) continue;
             mahjong[ting]++;
             hu();
             mahjong[ting]--;
         }
         if(!flag) printf(" Not ready");
         printf("\n");
     }
     return ;
 }

注：本随笔整理自QQ空间旧文。发布时间为2017年1月22日。

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