这道题我去年naq之前写过，但是没写出来，因为要记录路径什么的，当时的我啥也不会，今年又要naq了，我这次上来仔细重读了一下题，不就是个弱化版的记录路径spfa么，这有何难？

记录路径，费用流吼啊，行输入怎么办？stringstream吼啊

我说怎么这么多队过，原来是大水题hhh

#include <bits/stdc++.h>
#include <bits/extc++.h>
using namespace std;
#define limit (200000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x) {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int n,m,vs,ve;
int head[limit],cnt;
int dist[limit],fa[limit],vis[limit],pre[limit];
struct node{
    int to, next, flow, w;
}edge[limit<<1];
void add_one(int u, int v, int flow, int w){
    edge[cnt].to = v;
    edge[cnt].flow = flow;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
void add(int u, int v, int flow ,int w){
    add_one(u,v,flow,w);
    add_one(v,u,0,-w);
}
void init(int flag = 0){
    if(flag){
        memset(dist, INF, sizeof(dist));
        memset(vis, 0 , sizeof(vis));
    }else{
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
}
int spfa(){
    queue<int>q;
    init(1);
    dist[vs] = 0;
    vis[vs] = 1;
    q.push(vs);
    pre[ve] = -1;
    while (q.size()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            int stream = edge[i].flow;
            if(dist[u] + w < dist[v] && stream > 0){
                dist[v] = dist[u] + w;
                pre[v] = i;
                fa[v] = u;
                if(!vis[v]){
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return ~pre[ve];
}

int max_flow,min_cost;
map<int,string>mp;
map<string, int>id;
stack<string>stk;
void dinic(){
    if(!spfa()){
        cout<<"no route found"<<endl;
    }else{
        for(int i = ve; i != vs; i = fa[i]){
            stk.push(mp[i]);
        }
        stk.push(mp[vs]);
        while (stk.size()){
            cout<<stk.top()<<' ';
            stk.pop();
        }
    }
}
void solve(){
    cin>>n;
    init();
    string str;
    getline(cin,str);
    int tot = 0;
    rep(_,1,n){
        getline(cin,str);
        stringstream input(str);
        string st;
        input>>st;
        //cout<<st<<endl;
        if(!id[st])id[st] = ++tot,mp[tot]= st;
        string t;
        
        while (input>>t){
            if(!id[t])id[t] = ++tot,mp[tot]= t;
            add(id[st],id[t],1,0);
            add(id[t],id[st],1,0);
        }
    }
    string s,t;
    cin>>s>>t;
    vs = id[s],ve = id[t];
    dinic();
}
int main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    FASTIO
    //cin>>kase;
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}