Codeforces Round #566 (Div. 2) C. Beautiful Lyrics

鏈接：

https://codeforces.com/contest/1182/problem/C

題意：

You are given n words, each of which consists of lowercase alphabet letters. Each word contains at least one vowel. You are going to choose some of the given words and make as many beautiful lyrics as possible.

Each lyric consists of two lines. Each line consists of two words separated by whitespace.

A lyric is beautiful if and only if it satisfies all conditions below.

The number of vowels in the first word of the first line is the same as the number of vowels in the first word of the second line.
The number of vowels in the second word of the first line is the same as the number of vowels in the second word of the second line.
The last vowel of the first line is the same as the last vowel of the second line. Note that there may be consonants after the vowel.
Also, letters "a", "e", "o", "i", and "u" are vowels. Note that "y" is never vowel.

For example of a beautiful lyric,

"hello hellooowww"
"whatsup yowowowow"

is a beautiful lyric because there are two vowels each in "hello" and "whatsup", four vowels each in "hellooowww" and "yowowowow" (keep in mind that "y" is not a vowel), and the last vowel of each line is "o".
For example of a not beautiful lyric,

"hey man"
"iam mcdic"

is not a beautiful lyric because "hey" and "iam" don't have same number of vowels and the last vowels of two lines are different ("a" in the first and "i" in the second).
How many beautiful lyrics can you write from given words? Note that you cannot use a word more times than it is given to you. For example, if a word is given three times, you can use it at most three times.

思路：

模擬，記錄元音個數和最后一個元音，根據個數，和最后一個元音排序。將元音個數相等最后一個元音不等的放到一個對里，將個數相等最后一個元音也相等的放到另一個對里。
挨個輸出。當元音相等的較多時，補充一下即可。
因為vector的size是無符號整數，不能直接相減，因為這個wa2多次。。

代碼：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;

struct Word
{
    string word;
    int num;
    char last;
    bool operator < (const Word& that) const
    {
        if (this->num != that.num)
            return this->num < that.num;
        return this->last < that.last;
    }
}words[MAXN];

int main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> n;
    for (int i = 1;i <= n;i++)
    {
        cin >> words[i].word;
        for (auto c:words[i].word)
        {
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
            {
                words[i].num++;
                words[i].last = c;
            }
        }
    }
    sort(words+1, words+1+n);
    vector<pair<int, int> > fi, se;
    int pos = 1;
    int w = -1, cnt = 1;
    while(pos <= n)
    {
        if (words[pos].num == words[pos+1].num && words[pos].last == words[pos+1].last)
        {
            se.emplace_back(make_pair(pos, pos+1));
            pos += 2;
        }
        else if (words[pos].num != cnt)
        {
            w = pos;
            cnt = words[pos].num;
            pos++;
        }
        else if (w == -1)
        {
            w = pos;
            pos++;
        }
        else
        {
            fi.emplace_back(make_pair(w, pos));
            w = -1;
            pos++;
        }
    }
    int s1 = fi.size(), s2 = se.size();
    int res = 0;
    res += min(s1, s2) + max(0, (s2-s1)/2);
    cout << res << endl;
    int i;
    for (i = 0;i < min(fi.size(), se.size());i++)
    {
        cout << words[fi[i].first].word << ' ' << words[se[i].first].word << endl;
        cout << words[fi[i].second].word << ' ' << words[se[i].second].word << endl;
    }
    for (;i+1 < se.size();i+=2)
    {
        cout << words[se[i].first].word << ' ' << words[se[i+1].first].word << endl;
        cout << words[se[i].second].word << ' ' << words[se[i+1].second].word << endl;
    }

    return 0;
}