字符串搜索。HOJ1530 Compound Words。
2011-06-09 20:41
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stl set实现字符串搜索。。效率一般。(附二分搜索。)
Compound Words
Source:
Waterloo ACM Programming Contest Sep 28, 1996
You are to find all the two-word compound words in a dictionary. A two-word compound
word is a word in the dictionary that is the concatenation of exactly two other
words in the dictionary.
Input
Standard input consists of a number of lowercase words, one per line, in alphabetical
order. There will be no more than 120,000 words.
Output
Your output should contain all the compound words, one per line, in alphabetical
order.
Sample Input
Sample Output
二分实现。效率高很多。。
View Code
[/code]
Compound Words
Time limit: | 1sec. | Submitted: | 233 |
Memory limit: | 32M | Accepted: | 81 |
Waterloo ACM Programming Contest Sep 28, 1996
You are to find all the two-word compound words in a dictionary. A two-word compound
word is a word in the dictionary that is the concatenation of exactly two other
words in the dictionary.
Input
Standard input consists of a number of lowercase words, one per line, in alphabetical
order. There will be no more than 120,000 words.
Output
Your output should contain all the compound words, one per line, in alphabetical
order.
Sample Input
a alien born less lien never nevertheless new newborn the zebra
Sample Output
alien newborn 说明: 在给出的词中找出可有有任意另外两个词拼接而成的词输出。 用set存储与搜索。 代码如下: [code]#include<iostream> #include<set> using namespace std; int main() { set<string> s; string tmp; while (cin >> tmp) s.insert(tmp); set<string>::iterator it = s.begin(); for (it; it != s.end(); it++) { tmp = *it; for (int i = 1; i < tmp.length(); i++) { if (s.find(tmp.substr(0, i)) != s.end() && s.find(tmp.substr(i, tmp.length() - i)) != s.end()) { cout << tmp << endl; break; } } } return 0; }
二分实现。效率高很多。。
View Code
#include<stdio.h> #include<string.h> char ss[120010][50]; bool Bsearch(char *s, int n); int main() { char s1[100], *s2; int i = 0, j, k, n, len; while (scanf("%s", ss[i]) != EOF) i++; n = i; for (i = 0; i < n; i++) { len = strlen(ss[i]); for (j = 0, k = 0; j < len - 1; j++) { s1[k++] = ss[i][j]; s1[k] = '\0'; s2 = ss[i] + j + 1; if (Bsearch(s1, n) && Bsearch(s2, n)) { puts(ss[i]); break; } } } return 0; } bool Bsearch(char *s, int n) { int left = 0, right = n - 1, middle, r; while (left <= right) { middle = (left + right) / 2; r = strcmp(s, ss[middle]); if (r == 0) return true; if (r < 0) right = middle - 1; else left = middle + 1; } return false; }
[/code]
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