一.题目链接：

ZOJ-3700

二.题目大意：

T 组数据.

n 行字符串.

现有操作：将每个单词按照出现次数分组，同组中先按长度从大到小，长度相等时，再按字典序排列.

输出规则：在出现次数 > 1 的组，输出里面最长的单词，如果最长的单词不唯一，则输出最长单词中字典序排倒数第二的字符串.

三.分析：

水题！！！

申请两个 map，一个放每个单词出现次数，另一个放出现一定次数的所有单词（优先队列呀！）.

注意：".abc."，这种的要把 "abc" 分离出来（无用字符换成空格用 stringstream 分割即可），缩写算作一个单词.

应该 1A 的，5555. I am so vegetable.

四.代码实现：

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-6
#define PI acos(-1.0)
#define ll long long int
using namespace std;

struct cmp
{
    bool operator()(string s1, string s2)
    {
        if(s1.size() == s2.size())
            return s1 < s2;
        return s1.size() < s2.size();
    }
};

bool check(char s)
{
    if(s >= 'a' && s <= 'z' || s == '\'' || s == ' ')
        return 1;
    return 0;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        getchar();
        map <string, int> cnt;
        map <int, priority_queue <string, vector <string>, cmp> > mp;
        vector <string> ans;
        while(n--)
        {
            string str;
            getline(cin, str);
            int len = str.size();
            for(int i = 0; i < len; ++i)
            {
                if(str[i] >= 'A' && str[i] <= 'Z')
                    str[i] += 32;
                if(!check(str[i]))
                    str[i] = ' ';
            }
            stringstream ss;
            ss << str;
            while(ss >> str)
                cnt[str]++;
        }
        for(map <string, int> ::iterator iter = cnt.begin(); iter != cnt.end(); ++iter)
            if(iter->second > 1)
                mp[iter->second].push(iter->first);
        for(map <int, priority_queue <string, vector <string>, cmp> > ::iterator iter = mp.begin(); iter != mp.end(); ++iter)
        {
            priority_queue <string, vector <string>, cmp> q;
            q = iter->second;
            if(q.size() == 1)
                ans.push_back(q.top());
            else
            {
                string s1, s2;
                s1 = q.top();
                q.pop();
                s2 = q.top();
                if(s1.size() == s2.size())
                    ans.push_back(s2);
                else
                    ans.push_back(s1);
            }
        }
        int len = ans.size();
        bool flag = 0;
        for(int i = len - 1; i >= 0; --i)
        {
            if(flag)
                printf(" ");
            cout << ans[i];
            flag = 1;
        }
        printf("\n");
    }
    return 0;
}