xtu字符串 A. Babelfish
字符串
2023-09-11 14:15:29 时间
A. Babelfish
Time Limit: 3000ms
Memory Limit: 65536KB
64-bit integer IO format: %lld Java class name: MainYou have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.
Input
Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.
Output
Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".
Sample Input
dog ogday cat atcay pig igpay froot ootfray loops oopslay atcay ittenkay oopslay
Sample Output
cat eh loops
Hint
Huge input and output,scanf and printf are recommended.
解题:可以用trie,不过刚学trie,写起来貌似代价很大,而且由于是先读完数据,再查询的,不是动态修改查询的,这种查询折半查找就能完成。
先上TLE代码,本来以为STL可以完成的,没想到超时
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #include <map> 10 #define LL long long 11 #define INF 0x3f3f3f 12 using namespace std; 13 map<string,string>dic; 14 char str[200],a[100],b[100],word[100]; 15 int main(){ 16 while(gets(str) && str[0] != '\0'){ 17 sscanf(str,"%s %s",a,b); 18 dic.insert(pair<string,string>(b,a)); 19 } 20 map<string,string>::iterator it; 21 while(gets(word)){ 22 if(dic.count(word)) { 23 it = dic.find(word); 24 printf("%s\n",it->second.c_str()); 25 }else puts("eh"); 26 } 27 return 0; 28 }
下面是折半查找的代码,速度很快,代码很短,很喜欢呀。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #include <map> 10 #define LL long long 11 #define INF 0x3f3f3f 12 using namespace std; 13 struct word { 14 char from[20],to[20]; 15 } dic[100010]; 16 bool cmp(const word &a,const word &b) { 17 if(strcmp(a.from,b.from) <= 0) 18 return true; 19 return false; 20 } 21 int bsearch(int lt,int rt,char *str) { 22 int mid; 23 while(lt <= rt) { 24 mid = (lt+rt)>>1; 25 if(strcmp(dic[mid].from,str) == 0) return mid; 26 if(strcmp(str,dic[mid].from) < 0) rt = mid-1; 27 else lt = mid+1; 28 } 29 return -1; 30 } 31 int main() { 32 int cnt = 0,i,ans; 33 char str[50]; 34 while(gets(str) && str[0] != '\0') { 35 sscanf(str,"%s %s",dic[cnt].to,dic[cnt].from); 36 cnt++; 37 } 38 sort(dic,dic+cnt,cmp); 39 while(gets(str)) { 40 ans = bsearch(0,cnt,str); 41 if(ans == -1) puts("eh"); 42 else printf("%s\n",dic[ans].to); 43 } 44 return 0; 45 }
好吧,补上Trie的代码,trie更快啊!叼得不行。。。。。。。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #define LL long long 13 #define INF 0x3f3f3f3f 14 using namespace std; 15 const int maxn = 100010; 16 struct trie { 17 int letter[27]; 18 bool flag[27]; 19 int index[27]; 20 }; 21 trie dic[maxn]; 22 char to[maxn][20]; 23 int tot,cnt,root; 24 char str[50],a[20],b[20]; 25 void insertWord(int &root,int cur,const int len,char *s) { 26 if(dic[root].letter[s[cur]-'a'] == 0) 27 dic[root].letter[s[cur]-'a'] = tot++; 28 if(cur == len-1) { 29 dic[root].index[s[cur]-'a'] = cnt; 30 dic[root].flag[s[cur]-'a'] = true; 31 return; 32 } 33 insertWord(dic[root].letter[s[cur]-'a'],cur+1,len,s); 34 } 35 int query(int root,int cur,const int len,char *s) { 36 if(cur == len-1) { 37 if(dic[root].flag[s[cur]-'a']) 38 return dic[root].index[s[cur]-'a']; 39 return -1; 40 } 41 if(dic[root].letter[s[cur]-'a']) { 42 int v = dic[root].letter[s[cur]-'a']; 43 return query(v,cur+1,len,s); 44 } 45 return -1; 46 } 47 int main() { 48 tot = 1; 49 cnt = root = 0; 50 memset(dic,0,sizeof(dic)); 51 while(gets(str)&&str[0]!='\0') { 52 sscanf(str,"%s %s",a,b); 53 strcpy(to[cnt],a); 54 insertWord(root,0,strlen(b),b); 55 cnt++; 56 } 57 while(gets(str)) { 58 int ans = query(0,0,strlen(str),str); 59 if(ans == -1) puts("eh"); 60 else puts(to[ans]); 61 } 62 return 0; 63 }
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