航空路线问题

来回走就把它当成走两条既可以了
费用流跑一跑

# include 
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(210), __(1e6 + 10), INF(2e9);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int n, m, ans[_], num, mark[__];
map <string, int> M;
string name[_];
int cnt, fst[_], w[__], to[__], nxt[__], dis[_], vis[_], S, T, cost[__], pe[_], pv[_], max_flow, max_cost;
queue <int> Q;

IL void Add(RG int u, RG int v, RG int f, RG int co){
    cost[cnt] = co; w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
    cost[cnt] = -co; w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}

IL bool Bfs(){
    Q.push(S); Fill(dis, 127); dis[S] = 0; vis[S] = 1;
    while(!Q.empty()){
        RG int u = Q.front(); Q.pop();
        for(RG int e = fst[u]; e != -1; e = nxt[e]){
            if(!w[e] || dis[to[e]] <= dis[u] + cost[e]) continue;
            dis[to[e]] = dis[u] + cost[e];
            pe[to[e]] = e; pv[to[e]] = u;
            if(!vis[to[e]]) vis[to[e]] = 1, Q.push(to[e]);
        }
        vis[u] = 0;
    }
    if(dis[T] >= dis[2 * n + 1]) return 0;
    RG int ret = INF;
    for(RG int u = T; u != S; u = pv[u]) ret = min(ret, w[pe[u]]);
    for(RG int u = T; u != S; u = pv[u]) w[pe[u]] -= ret, w[pe[u] ^ 1] += ret;
    max_cost -= ret * dis[T]; max_flow += ret;
    return 1;
}

IL void Out(RG int u){
    ans[++num] = u;
    if(u == n) return;
    for(RG int e = fst[u + n]; e != -1; e = nxt[e])
        if(!mark[e] && w[e ^ 1]){  mark[e] = 1; Out(to[e]); return;  }
}

int main(RG int argc, RG char *argv[]){
    Fill(fst, -1); n = Read(); m = Read();
    Add(S, 1, 2, 0); T = n;
    for(RG int i = 1; i <= n; ++i) cin >> name[i], M[name[i]] = i;
    for(RG int i = 2; i < n; ++i) Add(i, i + n, 1, -1);
    Add(1, n + 1, 2, 0);
    for(RG int i = 1; i <= m; ++i){
        cin >> name[n + 1] >> name[n + 2];
        RG int u = M[name[n + 1]], v = M[name[n + 2]];
        if(u > v) swap(u, v);
        Add(u + n, v, INF, 0);
    }
    while(Bfs());
    if(max_flow != 2){  puts("No Solution!"); return 0;  }
    printf("%d\n", max_cost + 2);
    Out(1);
    for(RG int i = 1; i <= num; ++i) if(ans[i] && ans[i] <= n) cout << name[ans[i]] << endl;
    num = 0; Out(1);
    for(RG int i = num; i >= 1; --i) if(ans[i] != n && ans[i] <= n) cout << name[ans[i]] << endl;
    return 0;
}