http://acm.timus.ru/problem.aspx?space=1&num=1137
任何一个线路都是一个环 任意两个环如果有至少一个共同的节点 则两个环可以扩展成一个大的环
只要用dfs搜一遍记录路径就可以了
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#define LL long long
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
//const int N=105;
const int M=1005;
const int L=10005;
vector<int>str[L];
vector<bool>visited[L];
vector<int>ans;
int a[M];
void dfs(int x,int k)
{
for(unsigned int i=0;i<str[x].size();++i)
{
if(!visited[x][i])
{
visited[x][i]=true;
ans.insert(ans.begin()+k,str[x][i]);
dfs(str[x][i],k+1);
}
}
}
int main()
{
//freopen("data.txt","r",stdin);
int n;
while(cin>>n)
{
int m;
for(int i=1;i<L;++i)
{str[i].clear();visited[i].clear();}
int num=0;
int st=-1;
while(n--)
{
cin>>m;
num+=m;
for(int i=0;i<=m;++i)
{
cin>>a[i];
if(st==-1)
st=a[i];
if(i)
{
str[a[i-1]].push_back(a[i]);
visited[a[i-1]].push_back(false);
}
}
}
ans.clear();
ans.push_back(st);
dfs(st,1);
if(num!=ans.size()-1)
cout<<"0"<<endl;
else
{
cout<<num;
for(int i=0;i<ans.size();++i)
cout<<" "<<ans[i];
cout<<endl;
}
}
return 0;
}
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