題意：輸入n種牌以及牌的屬性，任意選3張，這3張滿足4種屬性，要麼全相同，要麼全不同，“*”是萬能牌，可以變成你想要的任意的牌，輸出3張拍的序號。

題解：暴力列舉

 1 //暴力
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<string>
 7 #include<unordered_map>
 8 #include<set>
 9 using
 
 namespace std;
10 
11 const int N = 260;
12 
13 char s[N][50];
14 string ss[N][6];
15 
16 
17 bool worng(string fir, string sec, string thi) {
18     if (fir == "*" || sec == "*" || thi == "*")   return 0; //只要有一個為萬能*則符合
19     else {
20         if (fir == sec && sec == thi)   return 0;
21         if
 
 (fir != sec && fir != thi && sec != thi) return 0;
22     }
23     return 1;
24 }
25 
26 bool judge(int id1, int id2, int id3) {
27     bool one = worng(ss[id1][1],ss[id2][1],ss[id3][1]);
28     bool two = worng(ss[id1][2],ss[id2][2],ss[id3][2]);
29     bool three = worng(ss[id1][3],ss[id2][3
 
],ss[id3][3]);
30     bool four = worng(ss[id1][4],ss[id2][4],ss[id3][4]);
31     if (one || two || three || four)    return false;
32     return true;
33 }
34 
35 void solve(int n, int t) {
36     for (int i = 0; i < n - 2; i++) {
37         for (int j = i+1; j < n - 1; j++) {
38             for (int k = j+1; k < n; k++) {
39                 if (judge(i, j, k)) {
40                     printf("Case #%d: %d %d %d\n", t, i+1, j+1, k+1);
41                     return;
42                 }
43             }
44         }
45     }
46     printf("Case #%d: -1\n", t);
47 }
48 
49 int main() {
50     int T;
51     cin >> T;
52     for (int cas = 1; cas <= T; cas++) {
53         int n;
54         cin >> n;
55         for (int i=0; i<n; i++){
56             scanf ("%s",s[i]);
57             int len=strlen(s[i]);
58             int cnt=1;
59             ss[i][1]=ss[i][2]=ss[i][3]=ss[i][4]="";
60             for (int j=0; j<len; j++){
61                 if (s[i][j]==']') {cnt++; continue;}
62                 else if (s[i][j]=='[') continue;
63                 ss[i][cnt]+=s[i][j];
64             }
65         }
66         solve(n, cas);
67     }
68     return 0;
69 }