果然正则表达式是一个强大的工具
更短的代码....hhh
版本1: 正则表达式..so easy~~
1 #include <iostream> 2 #include <algorithm> 3 #include <vector> 4 #include <string> 5 #include <regex> 6 using namespace std; 7 const int N = 107; 8 const string s_int = "([0-9]+)"; 9 const string s_str = "([0-9A-Za-z-_.]+)"; 10 const string s_path = "([0-9A-Za-z-_./]+)"; 11 string rule[N], na[N]; 12 vector <int> p[N]; // 表示匹配项类型 1-int;2-str;3-path 13 int n, m; 14 smatch sm; 15 string _deal(string str, int k) {// 将"int"->s_int'... 变成正则表达式的情况 16 string ans; 17 for (int i = 0; i < str.size(); i++) { 18 if (str[i] == '<') { 19 string tmp; i++; 20 while (str[i] != '>') 21 tmp.push_back(str[i++]); 22 if (tmp == "int") { ans += s_int; p[k].push_back(1); } 23 else if (tmp == "str") { ans += s_str; p[k].push_back(2); } 24 else { ans += s_path; p[k].push_back(3); } 25 } 26 else ans.push_back(str[i]); 27 } 28 return ans; 29 } 30 int main() 31 { 32 cin >> n >> m; 33 for (int i = 1; i <= n; i++) { 34 string s1; 35 cin >> s1 >> na[i]; 36 rule[i] = _deal(s1, i); 37 } 38 for (int i = 1; i <= m; i++) { 39 string str; cin >> str; 40 string ans = "404"; int k = 1; 41 for (; k <= n; k++) { 42 regex re(rule[k]); 43 if (regex_match(str,sm,re)) { 44 ans = na[k]; 45 break; 46 } 47 } 48 cout << ans; 49 if (ans != "404") 50 for (int j = 0; j < p[k].size(); j++) { 51 if (p[k][j] == 1) cout << " " <<stoi(sm[j+1]); 52 else cout << " " << sm[j+1]; 53 } 54 cout << " "; 55 } 56 //system("pause"); 57 return 0; 58
版本2: split() 用/来分割字符串 [以前写的 好复杂啊
1 #include <bits/stdc++.h> 2 using namespace std; 3 vector <string> name(107); 4 vector < vector <string> > rule(107); 5 vector <string> ans; 6 bool wei[107]; 7 bool isdight (char x) { 8 return x>='0'&&x<='9'; 9 } 10 bool ischar (char x) { 11 if (isdight(x) ) 12 return true; 13 if ( (x>='A' && x<='Z') || (x>='a' && x<='z') ) 14 return true; 15 if ( x=='/' || x=='-' || x=='_' || x=='.') 16 return true; 17 return false; 18 } 19 vector <string> split (const string& str,const char flag=' ') { 20 istringstream iss(str); 21 vector <string> sv; 22 string tmp; 23 while (getline(iss,tmp,flag)) 24 sv.push_back(tmp); 25 return sv; 26 } 27 bool isok (string str) { 28 if (str[0]!='/') return 0; 29 for (int i=0;i<str.size();i++) { 30 if (!ischar(str[i])) 31 return 0; 32 } 33 return 1; 34 35 } 36 string dtoc (int x) { 37 string str; 38 if (x==0) return "0"; 39 while (x) { 40 str+=x%10+'0'; 41 x/=10; 42 } 43 reverse(str.begin(),str.end()); 44 return str; 45 } 46 bool ismatch (int k, string str) { 47 bool flag=0; 48 ans.clear(); 49 if (str[str.size()-1]=='/') flag=1; 50 vector <string> sv=split (str,'/'); 51 52 for (int i=0;i<rule[k].size();i++) { 53 if (i>=sv.size()) return 0; 54 int len=sv[i].size(); 55 if (rule[k][i]=="<int>") { 56 int sum=0; 57 for (int j=0;j<len;j++) { 58 if (!isdight(sv[i][j])) return 0; 59 sum=sum*10+sv[i][j]-'0'; 60 } 61 ans.push_back(dtoc(sum)); 62 } 63 64 else if (rule[k][i]=="<str>") 65 ans.push_back(sv[i]); 66 67 else if (rule[k][i]=="<path>") { 68 string tmp; 69 for (int k=i;k<sv.size();k++) { 70 tmp+=sv[k]; 71 if (k==sv.size()-1&&!flag) continue; 72 tmp+="/"; 73 } 74 ans.push_back(tmp); 75 return 1; 76 } 77 78 else { 79 if (sv[i]!=rule[k][i]) return 0; 80 } 81 82 } 83 if (rule[k].size()==sv.size()&&(flag==wei[k])) return 1; // 真的是傻逼 84 else return 0; 85 } 86 int n,m; 87 int main () 88 { 89 cin>>n>>m; getchar(); 90 string s1,s2; 91 for (int i=1;i<=n;i++) { 92 cin>>s1>>s2; getchar(); 93 name[i]=s2; 94 if (s1[s1.size()-1]=='/') wei[i]=1;// 没有考虑的地方 95 rule[i]=split(s1,'/'); 96 } 97 for (int i=1;i<=m;i++) { 98 getline (cin,s1); int x=0; 99 if (isok(s1)) { 100 for (int j=1;j<=n;j++) 101 if (ismatch(j,s1)) { 102 x=j; 103 break; 104 } 105 } 106 if (x==0) cout<<"404"<<endl; 107 else { 108 cout<<name[x]; 109 for (int j=0;j<ans.size();j++) 110 cout<<" "<<ans[j]; 111 cout<<endl; 112 } 113 } 114 return 0; 115 }