首先预处理,用g数组来存储任意两个单词之间最长的重合部分,used数组来记录每个单词使用的次数,然后就是即可暴搜。
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i = a;i<n;i++)
#define per(i,a,n) for(int i = n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define yes cout<<"YES"<<'\n';
#define no cout<<"NO"<<'\n';
#define endl '\n';
#define R register
typedef vector<int> VI;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll MOD=1000000007;
int rnd(int x) {return mrand() % x;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;};
ll lcm(int a,int b){return a*b/gcd(a,b);};
int n;
string s[25];
int g[25][25];
int used[25];
int ans;
void dfs(string t,int last){
ans=max(SZ(t),ans);
used[last]++;
rep(i,0,n){
if(g[last][i]&&used[i]<2) dfs(t+s[i].substr(g[last][i]),i);
}
used[last]--;
}
int main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin>>n;
rep(i,0,n) cin>>s[i];
char st;
cin>>st;
rep(i,0,n){
rep(j,0,n){
auto a=s[i],b=s[j];
rep(k,1,min(SZ(a),SZ(b))){
if(a.substr(SZ(a)-k)==b.substr(0,k)){
g[i][j]=k;
break;
}
}
}
}
rep(i,0,n){
if(s[i][0]==st) dfs(s[i],i);
}
cout<<ans<<endl;
return 0;
}
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