题意：输入n种牌以及牌的属性，任意选3张，这3张满足4种属性，要么全相同，要么全不同，“*”是万能牌，可以变成你想要的任意的牌，输出3张拍的序号。
题解：暴力枚举
更新一种更简洁明了的代码
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 vector <string> vet;
13 unordered_map<string, string> ma, mb, mc, md, mp;
14
15 void init() {
16 mp["one"]="1";
17 mp["two"]="2";
18 mp["three"]="3";
19 mp["diamond"]="1";
20 mp["squiggle"]="2";
21 mp["oval"]="3";
22 mp["solid"]="1";
23 mp["striped"]="2";
24 mp["open"]="3";
25 mp["red"]="1";
26 mp["green"]="2";
27 mp["purple"]="3";
28 mp["*"]="0";
29 }
30
31
32 inline string read() {
33 string p;
34 p = "";
35 char ch = getchar();
36 while ((ch < 'a' || ch > 'z') && ch != '*') {
37 ch = getchar();
38 }
39 while ((ch >= 'a' && ch <= 'z') || ch == '*') {
40 p += ch;
41 ch = getchar();
42 }
43 return p;
44 }
45
46 bool worng(char fir, char sec, char thi) {
47 if (fir == '0' || sec == '0' || thi == '0') return 0; //只要有一个为万能*则符合
48 else {
49 if (fir == sec && sec == thi) return 0;
50 if (fir != sec && fir != thi && sec != thi) return 0;
51 }
52
53 return 1;
54 }
55
56 bool judge(int i, int j, int k) {
57 string fri = vet[i], sec = vet[j], thr = vet[k];
58 bool one = worng(fri[0], sec[0], thr[0]);
59 bool two = worng(fri[1], sec[1], thr[1]);
60 bool three = worng(fri[2], sec[2], thr[2]);
61 bool four = worng(fri[3], sec[3], thr[3]);
62 if (one || two || three || four) return false;
63 return true;
64 }
65
66 void solve(int n, int t) {
67 for (int i = 0; i < n - 2; i++) {
68 for (int j = i+1; j < n - 1; j++) {
69 for (int k = j+1; k < n; k++) {
70 if (judge(i, j, k)) {
71 printf("Case #%d: %d %d %d
", t, i+1, j+1, k+1);
72 return;
73 }
74 }
75 }
76 }
77 printf("Case #%d: -1
", t);
78 }
79
80 int main() {
81 int T;
82 cin >> T;
83 init();
84 for (int cas = 1; cas <= T; cas++) {
85 int n;
86 vet.clear();
87 cin >> n;
88 for (int i = 0; i < n; i++) {
89 string one = read();
90 string two = read();
91 string three = read();
92 string four = read();
93 vet.push_back({mp[one]+mp[two]+mp[three]+mp[four]});
94 }
95 solve(n, cas);
96 }
97 return 0;
98 }
分割线——————————————————————
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; //只要有一个为万能*则符合
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
", t, i+1, j+1, k+1);
41 return;
42 }
43 }
44 }
45 }
46 printf("Case #%d: -1
", 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 }