题目：
思路分析：
最开始以为要建分层图可是一看数据范围
dfs暴力也可以呀
就是最短路+路径存储的题目 当然dijsktra也行！
代码实现：
/*
*@Author: GuoJinlong
*@Language: C++
*/
//#include <bits/stdc++.h>
/*
* __----~~~~~~~~~~~------___
* . . ~~//====...... __--~ ~~
* -. \_|// |||\\ ~~~~~~::::... /~
* ___-==_ _-~o~ \/ ||| \\ _/~~-
* __---~~~.==~||\=_ -_--~/_-~|- |\\ \\ _/~
* _-~~ .=~ | \\-_ '-~7 /- / || \ /
* .~ .~ | \\ -_ / /- / || \ /
* / ____ / | \\ ~-_/ /|- _/ .|| \ /
* |~~ ~~|--~~~~--_ \ ~==-/ | \~--===~~ .\
* ' ~-| /| |-~\~~ __--~~
* |-~~-_/ | | ~\_ _-~ /\
* / \ \__ \/~ \__
* _--~ _/ | .-~~____--~-/ ~~==.
* ((->/~ '.|||' -_| ~~-/ , . _||
* -_ ~\ ~~---l__i__i__i--~~_/
* _-~-__ ~) \--______________--~~
* //.-~~~-~_--~- |-------~~~~~~~~
* //.-~~~--\
* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
*
* 神兽保佑 永无BUG
*/
const int MAX=10010;
vector<int>path,tmp;
vector<int>v[MAX];
int line[MAX][MAX];
int n,m,k;
int count(vector<int>a){
int pre=0;
int cnt=-1;
for(int i=1;i<a.size();i++){
if(line[a[i-1]][a[i]]!=pre)
cnt++;
pre=line[a[i-1]][a[i]];
}
return cnt;
}
int vis[MAX];
int m1,m2;
void dfs(int st,int en,int cnt){
if(st==en){
if(cnt<m1||(m2>count(tmp)&&m1==cnt)){
path=tmp;
m1=cnt;
m2=count(tmp);
}
return;
}
for(int i=0;i<v[st].size();i++){
int u=v[st][i];
if(vis[u]) continue;
vis[u]=1;
tmp.push_back(u);
dfs(u,en,cnt+1);
vis[u]=0;
tmp.pop_back();
}
}
int main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>k;
int x;
int pre;
cin>>pre;
for(int j=2;j<=k;j++){
cin>>x;
v[x].push_back(pre);
v[pre].push_back(x);
line[x][pre]=i;
line[pre][x]=i;
pre=x;
}
}
cin>>m;
while (m--) {
int x,y;
cin>>x>>y;
tmp.clear();
tmp.push_back(x);
vis[x]=1;
m1=INF;
m2=INF;
dfs(x,y,0);
// cout<<m1<<" "<<m2<<endl;
vis[x]=0;
if(m1>=INF) {cout<<"Sorry, no line is available.\n";continue;}
cout<<m1<<endl;
int p1=x;
int p2=0;
for(int i=1;i<path.size();i++){
if(line[path[i-1]][path[i]]!=p2){
if(p2!=0){
printf("Go by the line of company #%d from %04d to %04d.\n", p2, p1, path[i-1]);
}
p1=path[i-1];
p2=line[path[i-1]][path[i]];
}
}
printf("Go by the line of company #%d from %04d to %04d.\n", p2, p1, y);
}
}