字符串的匹配与排序。HOJ1469 Anagram Groups。

字符串的排序。匹配。

用stl vector实现。

Anagram Groups

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Time limit:	1sec.	Submitted:	114
Memory limit:	32M	Accepted:	35

Source: University of Ulm Internal Contest 2000

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World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups.

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group. Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace. It is terminated by EOF.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

说明：
给出若干词。找出Anagram词数最多的五组。 
即一个词如果将字母换一些位置可以得到另一个词。这两个词就算一组。每组按字典序输出。
五组之间先按size。然后按第一个词的字典序输出。
重复的词只输出一次。组数不足五个则全部输出即可。
思路是 存下 每个词的原词 和 词的所有字母按字典序重新排序后得到的词。
比如  ate eat  存下原词 ate eat 和 aet aet 这样重新排序后的词。
比较时直接比较排序后的词。 如果一样。 则它们是一组。
注意几点。 
1. 先排序。 这样比较的时候一次n循环就可以分出所有组。
2.重复的时候size也要加一。就是说会出现 Group of size 4: word . 类似的情况。

代码如下：
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;

struct words {
    string w;
    string ord;

    bool operator<(const words & x)const {
        if (ord != x.ord)return ord < x.ord;
        return w < x.w;
    }
};

struct group {
    vector<words> wor;

    bool operator<(const group & x)const {
        if (wor.size() != x.wor.size())return wor.size() > x.wor.size();
        return wor[0].w < x.wor[0].w;
    }
};

int main() {
    vector<words> word;
    vector<group> rr;
    vector<words> res;
    words tmp;
    while (cin >> tmp.w) {
        tmp.ord = tmp.w;
        sort(tmp.ord.begin(), tmp.ord.end());
        word.push_back(tmp);
    }
    sort(word.begin(), word.end());

    res.push_back(word[0]);
    group tt;
    for (int i = 1; i < word.size(); i++) {
        if (word[i].ord == word[i - 1].ord);
        else {
            tt.wor = res;
            rr.push_back(tt);
            res.clear();
        }
        res.push_back(word[i]);
    }
    tt.wor = res;
    rr.push_back(tt);
    sort(rr.begin(), rr.end());
    for (int i = 0; i < rr.size() && i < 5; i++) {
        cout << "Group of size " << rr[i].wor.size() << ": ";
        cout << rr[i].wor[0].w << " ";
        for (int j = 1; j < rr[i].wor.size(); j++) {
            if (rr[i].wor[j].w != rr[i].wor[j - 1].w)
                cout << rr[i].wor[j].w << " ";
        }
        cout << ".\n";
    }
    return 0;
}