[Explanations] Air Routes

Air Routes

A point in two ways, but only choose one with, and directly connected to a point removed a capacity of 1 on the side of limited success.

The two points of a split point, one exit point a point, point to the OUT point even as a Cost \ (- inf \) edge, running cost flow, the final answer is
\ [- \ dfrac {cost} { inf} \]

Output program look like a casual get. Note that the border situation consider! ! ! (1- "n), (do not communicate)

Forced Summary: The demolition site into a state of two points, to the middle of a limiting edge, one can simulate a point of view has only two states, more states can be even.

Cost flow can not run a negative ring. spfa run what cost flow on what can run.

//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>


using namespace std;  typedef long long ll;
inline int qr(){
      register int ret=0,f=0;
      register char c=getchar();
      while(c<48||c>57)f|=c==45,c=getchar();
      while(c>=48&&c<=57) ret=ret*10+c-48,c=getchar();
      return f?-ret:ret;
}
struct E{
      int to,nx,w,c;
      E(){to=nx=w=c=0;}
      E(const int&a,const int&b,const int&CX,const int&d){to=a;nx=b;w=CX;c=d;}
}e[30005];
int head[404];
int cnt=1;
inline void add(const int&fr,const int&to,const int&w,const int&c,const bool&f){
      e[++cnt]=E(to,head[fr],w,c);
      head[fr]=cnt;
      //printf("fr=%d to=%d w=%d c=%d cnt=%d\n",fr,to,w,c,cnt);
      if(f)add(to,fr,0,-c,0);
}
const int maxn=505;
int d[maxn];
int fl[maxn];
int in[maxn];
int last[maxn];

int S,T,n,v,cost;
const int inf=0x3f3f3f3;
const int littleinf=1;

queue < int > q;
void mincost(){
      
      while(1){
        for(register int t=1;t<=T;++t) d[t]=inf,fl[t]=0,last[t]=0,in[t]=0;
        d[S]=0;fl[S]=inf;
        q.push(S);
        while(!q.empty()){
          register int now=q.front();
          q.pop();
          in[now]=0;
          //cout<<now<<endl;
          for(register int t=head[now];t;t=e[t].nx){
            //cout<<t<<endl;
            if(d[e[t].to]>e[t].c+d[now]&&e[t].w>0){
                  d[e[t].to]=e[t].c+d[now];
                  if(!in[e[t].to]) q.push(e[t].to);
                  in[e[t].to]=1;
                  fl[e[t].to]=min(e[t].w,fl[now]);
                  last[e[t].to]=t;
            }
          }
        }
        if(fl[T]<=0)return;
        cost-=fl[T]*d[T];
        for(register int t=T;t!=S;t=e[last[t]^1].to)
          e[last[t]].w-=fl[T],e[last[t]^1].w+=fl[T];
      }
}

vector < int > ve,VE;
map < string , int > name;
map < int , string > arc;
int main(){     
      freopen("airl.in","r",stdin);
      freopen("airl.out","w",stdout);
      n=qr();v=qr();
      for(register int t=1;t<=n;++t){
        string f;
        cin>>f;
        name[f]=t;
        arc[t]=f;
        add(t+1,t+n+1,1+(t==1||t==n),0,1);
      }
      
      for(register int t=1;t<=v;++t){
        string t1,t2;
        cin>>t1>>t2;
        int i1,i2;
        i1=name[t1];i2=name[t2];
        if(i1>i2)swap(i1,i2);
        add(i1+n+1,i2+1,1+(i1==1&&i2==n),-1,1);
      }
      
      S=1;
      T=n+n+2;
      add(n+n+1,T,2,0,1);
      add(S,2,2,0,1);
      mincost();
      if(e[cnt].w!=2) return puts("No Solution!"),0;
      printf("%d\n",cost);
      int now=1;
      ve.push_back(now);
      while(now!=n){
        for(register int t=head[now+n+1];t;t=e[t].nx){
          if(e[t^1].w&&e[t].to>1&&e[t].to<=n+1&&!in[e[t].to-1]){
            now=e[t].to-1;
            break;
          }
        }
        in[now]=1;
        ve.push_back(now);
      }
      in[n]=0;
      now=1;
      while(now!=n){
        int f=1;
        for(register int t=head[now+n+1];t;t=e[t].nx){
          if(e[t^1].w&&e[t].to>1&&e[t].to<=n+1&&!in[e[t].to-1]){
            now=e[t].to-1;
            f=0;
            break;
          }
        }
        if(now==n||f)break;
        in[now]=1;
        VE.push_back(now);
      }
      reverse(VE.begin(),VE.end());
      VE.push_back(1);
      
      for(vector< int >::iterator t=ve.begin();t!=ve.end();++t) cout<<arc[*t]<<endl;
      for(vector< int >::iterator t=VE.begin();t!=VE.end();++t) cout<<arc[*t]<<endl;
      return 0;
}