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

链接：

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

题意：

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：

模拟，记录元音个数和最后一个元音，根据个数，和最后一个元音排序。将元音个数相等最后一个元音不等的放到一个对里，将个数相等最后一个元音也相等的放到另一个对里。

挨个输出。当元音相等的较多时，补充一下即可。

因为vector的size是无符号整数，不能直接相减，因为这个wa2多次。。

代码：

#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;
}