stl set實現字符串搜索。。效率一般。(附二分搜索。)

Compound Words

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Time limit:	1sec.	Submitted:	233
Memory limit:	32M	Accepted:	81

Source: Waterloo ACM Programming Contest Sep 28, 1996

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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

說明：
在給出的詞中找出可有有任意另外兩個詞拼接而成的詞輸出。
用set存儲與搜索。
代碼如下：
#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;
}
二分實現。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;
}

轉載于:https://www.cnblogs.com/yangchenhao8/archive/2011/06/09/2076812.html

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