就从给定字母开始不断找能接在最后一个单词后面且使用次数小于两次的单词。

#include<iostream>
#include<cassert>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<string>
#include<iterator>
#include<cstdlib>
#include<vector>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define rep(i,f,t) for(int i = (f),_end_=(t); i <= _end_; ++i)
#define rep2(i,f,t) for(int i = (f),_end_=(t); i < _end_; ++i)
#define dep(i,f,t) for(int i = (f),_end_=(t); i >= _end_; --i)
#define dep2(i,f,t) for(int i = (f),_end_=(t); i > _end_; --i)
#define clr(c, x) memset(c, x, sizeof(c) )
typedef long long int64;
const int INF = 0x5f5f5f5f;
const double eps = 1e-8;//*****************************************************string s[22];
int n;
int vis[22];
bool ok(const string &st, int i, int j)                  //判断单词st能否从i位置开始将第j个单词拼接上
{if(s[j].size() < st.size()-i)return false;for(int k = 0; i < st.size(); ++i, ++k){if(st[i] != s[j][k])return false;}return true;
}
int dfs(int t,int len)
{int ans = len;++vis[t];                               //单词使用次数rep(j,1,n){if(vis[j] >= 2)continue;int jian = 0;for(string::iterator i = s[t].end()-1; i != s[t].begin(); --i)               //尽量从单词尾部开始拼接{++jian;                      //与新单词重合的长度if(ok( s[t], i-s[t].begin() , j)){int tmp = dfs(j,len+s[j].size()-jian);ans = max(ans,tmp);                        //记录最大长度break;}}}--vis[t];                  //还原该单词使用次数return ans;
}
int solve(char c)
{int ans = 0;rep(i,1,n){if(s[i][0]==c){int tmp = dfs(i,s[i].size());ans = max(ans,tmp);}}return ans;
}
int main()
{cin>>n;rep(i,1,n)cin>>s[i];char c;cin>>c;int ans = solve(c);cout<<ans<<endl;return 0;
}