OJ題號：UVa11210

思路：

首先字符串處理讀入手牌，str數組將手牌和數字對應，接下來搜索，先搜對子，如果搜過對子就不搜了。由於對子有且只有一個，可以在搜到以後直接跳出。同時註意一副麻將中每種牌只有4個，所以不能作為將。

 1 #include<cstdio>
 2 #include<string>
 3 #include<vector>
 4 #include<sstream>
 5 #include<iostream>
 6 using namespace std;
 7 vector<int> mahjong;
 
 8 const string str[35]={"","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"};
 9 void input(string line,vector<int> &ans) {
10     ans.clear();
11     for(int i=0
 
;i<35;i++) ans.push_back(0);
12     stringstream ss(line);
13     string temp;
14     for(int i=0;i<13;i++) {
15         ss>>temp;
16         for(int j=1;j<=34;j++) {
17             if(temp==str[j]) {
18                 ans[j]++;
19                 break;
20             }
21         }
22
 
     }
23 }
24 bool duizi=0,flag=0;
25 int ting;
26 bool hu() {
27     for(int i=1;i<=34;i++) {
28         if(mahjong[i]) {
29             goto Try;
30         }
31     }
32     cout<<" "+str[ting];
33     flag=1;
34     return 1;
35     Try:
36     if(!duizi) {
37         for(int i=1;i<=34;i++) {
38             if(mahjong[i]>=2) {
39                 mahjong[i]-=2;
40                 duizi=1;
41                 bool f=hu();
42                 duizi=0;
43                 mahjong[i]+=2;
44                 if(f) return 1;
45             }
46         }
47         return 0;
48     }
49     for(int i=1;i<=34;i++) {
50         if(mahjong[i]>=3) {
51             mahjong[i]-=3;
52             bool f=hu();
53             mahjong[i]+=3;
54             if(f) return 1;
55         }
56     }
57     for(int i=1;i<=27;i++) {
58         if(i==8||i==9||i==17||i==18||i==26||i==27) continue;
59         if(mahjong[i]&&mahjong[i+1]&&mahjong[i+2]) {
60             mahjong[i]--;
61             mahjong[i+1]--;
62             mahjong[i+2]--;
63             bool f=hu();
64             mahjong[i]++;
65             mahjong[i+1]++;
66             mahjong[i+2]++;
67             if(f) return 1;
68         }
69     }
70     return 0;
71 }
72 int main() {
73     int turn=0;
74     while(1) {
75         string s;
76         getline(cin,s);
77         if(s=="0") break;
78         turn++;
79         printf("Case %d:",turn);
80         mahjong.clear();
81         input(s,mahjong);
82         flag=0;
83         for(ting=1;ting<=34;ting++) {
84             if(mahjong[ting]==4) continue;
85             mahjong[ting]++;
86             hu();
87             mahjong[ting]--;
88         }
89         if(!flag) printf(" Not ready");
90         printf("\n");
91     }
92     return 0;
93 }

註：本隨筆整理自QQ空間舊文。發布時間為2017年1月22日。

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Chinese Mahjong