題意：按規則拼詞
思路：先按元音結尾排序取第二種詞，再按數目取第一種詞，貪心的先用第一種詞，如果不夠，就第二種詞可以當第一種用.
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=1e5+10;
struct node
{
string word;
int num;
char last;
}words[MAXN];
int cmp(node aa,node bb)
{
if(aa.num==bb.num)
return aa.last<bb.last;
return aa.num<bb.num;
}
vector<pair<int, int> > fi, se;
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
string s;
cin>>s;
words[i].word=s;
for(int j=0;j<s.size();j++)
{
if (s[j] == 'a' || s[j] == 'e' || s[j] == 'i' || s[j] == 'o' || s[j] == 'u')
{
words[i].num++;
words[i].last = s[j];
}
}
}
sort(words+1,words+1+n,cmp);
int pos=1;
int cnt=0;
int w=-1;
while(pos<=n)
{
if (words[pos].num == words[pos+1].num && words[pos].last == words[pos+1].last)
{
se.push_back({pos, pos+1});
pos += 2;
continue;
}
if(w==-1)// 如果不符合 那麼和前面的位置進行比較
{
w=pos;
cnt=words[pos].num;
pos++;
continue;
}
if(words[pos].num==cnt)// 和前位置元音數量一樣 放入數量一樣的容器裏
{
fi.push_back({w, pos});
w = -1;
pos++;
}
else// 否則更新位置 因爲按照元音數量排序的
{
w = pos;
cnt = words[pos].num;
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(s1,s2);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;
}