codeforces 589H - Tourist Guide（dfs）

H - Tourist Guide

Time Limit:2000MS     Memory Limit:524288KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 589H

Description

It is not that easy to create a tourist guide as one might expect. A good tourist guide should properly distribute flow of tourists across the country and maximize the revenue stream from the tourism. This is why there is a number of conditions to be met before a guide may be declared as an official tourist guide and approved by Ministry of Tourism.

Ministry of Tourism has created a list of k remarkable cities out of n cities in the country. Basically, it means that in order to conform to strict regulations and to be approved by the ministry a tourist guide should be represented as a set of routes between remarkable cities so that the following conditions are met:

• the first and the last city of every route are distinct remarkable cities,
• each remarkable city can be an endpoint of at most one route,
• there is no pair of routes which share a road.

Please note that a route may pass through a remarkable city. Revenue stream from the tourism highly depends on a number of routes included in a tourist guide so the task is to find out a set of routes conforming the rules of a tourist guide with a maximum number of routes included.

Input

The first line contains three integer numbers n,  m,  k(1 ≤ n ≤ 50000,  0 ≤ m ≤ 50000,  1 ≤ k ≤ n) — the number of cities in the country, the number of roads in the country and the number of remarkable cities correspondingly.

Each of the following m lines contains two integer numbers a_i and b_i(1 ≤ a_i, b_i ≤ n) — meaning that cities a_i and b_i are connected by a bidirectional road. It is guaranteed that a_i and b_i are distinct numbers and there is no more than one road between a pair of cities.

The last line contains k distinct integer numbers — a list of remarkable cities. All cities are numbered from 1 to n.

Output

The first line of the output should contain c — the number of routes in a tourist guide. The following c lines should contain one tourist route each. Every route should be printed in a form of "tv₁v₂ ... v_t + 1", where t is a number of roads in a route and v₁,  v₂, ..., v_t + 1 — cities listed in the order they are visited on the route.

If there are multiple answers print any of them.

Sample Input

Input

6 4 4
1 2
2 3
4 5
5 6
1 3 4 6

Output

2
2 1 2 3
2 4 5 6

Input

4 3 4
1 2
1 3
1 4
1 2 3 4

Output

2
1 1 2
2 3 1 4

我发现我蠢得连建图都不会建。。。。
这道题虽然思路貌似并不难。。。但是写不出。。。因为我竟然不会用vector....
简直sad...
于是照猫画虎。。。熟悉一下vector...还有pair....
熟悉下建图。。。。。。
嗯。。慢慢来。

/*************************************************************************
    > File Name: code/hust/20151025/H.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年10月27日 星期二 19时30分28秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
                 
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
#define pb push_back
using namespace std;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=5E4+7;

int n,m,k;
vector<int> adj[N];
vector<pair< pair<int,int> ,int> > ans;
int pa[N];
int kt[N];
bool spe[N];
void init()
{
    ms(spe,false);
    ms(pa,0);
    scanf("%d %d %d",&n,&m,&k);
    for ( int i = 0,u,v; i < m ; i++)
    {
    scanf("%d %d",&u,&v);
    adj[u].push_back(v);
    adj[v].push_back(u);
    }

    for ( int i = 0,u ; i < k ; i++)
    {
    scanf("%d",&u);
    spe[u] = true;
    }

}

void dfs( int u)
{
 //   cout<<"u:"<<u<<endl;
    vector<int> tmp;
    tmp.clear();

    for ( int j = 0 ; j < adj[u].size(); j++)
    {
    int v =adj[u][j];
    if (pa[v]==0)
    {
        pa[v] =  u;
        dfs(v);
        if (kt[v]) tmp.push_back(kt[v]);
    }
    }
    while (tmp.size()>1)
    {
    int x1 = tmp.back();tmp.pop_back();
    int x2 = tmp.back(); tmp.pop_back();
//    cout<<"x1:"<<x1<<" x2:"<<x2<<endl;
    ans.pb(make_pair(make_pair(x1,x2),u));
    }
    if (tmp.size()>0)
    {
    int x = tmp.back();tmp.pop_back();

//    cout<<"x:"<<x<<endl;
    if (spe[u])
    {
        ans.push_back(make_pair(make_pair(x,u),u));
    }
    else kt[u] = x;
    }
    else 
    {
    if (spe[u]) kt[u] =  u;
    }
}

void solve()
{
    for ( int i = 1 ; i <n+1 ; i++)
    if (!pa[i])
    {
        pa[i] = -1;
        dfs(i);
    }

    vector<int>p,q;
   // printf("%d\n",ans.size());
      cout<<ans.size()<<endl;
    for ( int i = 0 ; i < ans.size(); i++)
    {
    int u = ans[i].first.first;
    int v = ans[i].first.second;
    int r = ans[i].second;
//    cout<<"************************"<<endl;
//    cout<<"u:"<<u<<" v:"<<v<<" r:"<<r<<endl;
//    cout<<"************************"<<endl<<endl;

    p.clear();
    q.clear();
    while (u!=r)
    {
        p.push_back(u);
        u = pa[u];
    }

    while (v!=r)
    {
        q.push_back(v);
        v = pa[v];
    }

    //printf("%d ",p.size()+q.size());
    cout<<p.size()+q.size()<<" ";
    for ( int j = 0 ; j < p.size(); j++) printf("%d ",p[j]);
    printf("%d ",r);
    
    reverse(q.begin(),q.end());
    for ( int j = 0 ; j < q.size(); j++) printf("%d ",q[j]);
    puts("");    
    
    }
}
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif
   init();
   solve();
  
   
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}