UVaOJ 156 - Ananagrams

问题描述

解法一：映射（Map）

使用 map<string, int> cnt 统计标准化 standard() 表示后单词 st_word 的出现数量，如果只出现过一次，说明该单词无法通过其它单词重新排序得到。注意输出也需要按照字典序。

#include <bits/stdc++.h>

using namespace std;

string standard(string &s) {
  string res = "";
  for (auto c:s) res += tolower(c);
  sort(res.begin(), res.end());
  return res;
}

int main() {
  string word, st_word;
  vector<string> words;
  map<string, int> cnt;
  while (cin >> word && word[0] != '#') {
    words.push_back(word);
    st_word = standard(word);
    if (cnt.find(st_word) == cnt.end()) cnt[st_word] = 0;
    cnt[st_word] += 1; 
  }
  sort(words.begin(), words.end());
  for (auto word: words) {
    st_word = standard(word);
    if (cnt[st_word] == 1) cout << word << endl;
  }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;

string standard(string &s) {

string res = "";

for (auto c:s) res += tolower(c);

sort(res.begin(), res.end());

return res;

}

int main() {

string word, st_word;

vector<string> words;

map<string, int> cnt;

while (cin >> word && word[0] != '#') {

words.push_back(word);

st_word = standard(word);

if (cnt.find(st_word) == cnt.end()) cnt[st_word] = 0;

cnt[st_word] += 1;

}

sort(words.begin(), words.end());

for (auto word: words) {

st_word = standard(word);

if (cnt[st_word] == 1) cout << word << endl;

}

return 0;

}

#TODO

#TODO

AOAPC II UVaOJ