poj--1904--King's Quest（scc建图）

King's Quest

Time Limit: 15000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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Description

Once upon a time there lived a king and he had N sons. And there were N beautiful girls in the kingdom and the king knew about each of his sons which of those girls he did like. The sons of the king were young and light-headed,
so it was possible for one son to like several girls.



So the king asked his wizard to find for each of his sons the girl he liked, so that he could marry her. And the king's wizard did it -- for each son the girl that he could marry was chosen, so that he liked this girl and, of course, each beautiful girl had
to marry only one of the king's sons.



However, the king looked at the list and said: "I like the list you have made, but I am not completely satisfied. For each son I would like to know all the girls that he can marry. Of course, after he marries any of those girls, for each other son you must
still be able to choose the girl he likes to marry."



The problem the king wanted the wizard to solve had become too hard for him. You must save wizard's head by solving this problem.

Input

The first line of the input contains N -- the number of king's sons (1 <= N <= 2000). Next N lines for each of king's sons contain the list of the girls he likes: first Ki -- the number of those girls, and then Ki different integer
numbers, ranging from 1 to N denoting the girls. The sum of all Ki does not exceed 200000.




The last line of the case contains the original list the wizard had made -- N different integer numbers: for each son the number of the girl he would marry in compliance with this list. It is guaranteed that the list is correct, that is, each son likes the
girl he must marry according to this list.

Output

Output N lines.For each king's son first print Li -- the number of different girls he likes and can marry so that after his marriage it is possible to marry each of the other king's sons. After that print Li different integer numbers
denoting those girls, in ascending order.

Sample Input

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

Sample Output

2 1 2
2 1 2
1 3
1 4

Hint

This problem has huge input and output data,use scanf() and printf() instead of cin and cout to read data to avoid time limit exceed.



有n个王子n个女生，输入时第一个数字是n，然后n行，每一行第一个数字表示该王子喜欢几个人，然后接这几个女孩的编号，第n+1行表示第i个女孩喜欢ni王子



将女孩虚拟为n+i节点，建图时i--j+n表示i王子喜欢女孩j，因为男女编号有重复，所以j+n，j+n--i表示j喜欢i，找出每一个scc里的点判断就好

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;
#define MAX 5000
int low[MAX],dfn[MAX];
int head[MAX],sccno[MAX],scc_cnt,dfs_clock;
int cnt,m,n,map[MAX][MAX],rec[MAX];
bool Instack[MAX];
stack<int>s;
vector<int>scc[MAX];
struct node
{
	int u,v;
	int next;
}edge[MAX*100];
void init()
{
	cnt=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void getmap()
{
	int num,a;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&num);
		while(num--)
		{
			scanf("%d",&a);
			add(i,a+n);
			map[i][a]=1;
		}
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a);
		add(a+n,i);
	}
}
void tarjan(int u,int fa)
{
	int v;
	low[u]=dfn[u]=++dfs_clock;
	s.push(u);
	Instack[u]=true;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].v;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
		}
		else if(Instack[v])
		low[u]=min(low[u],dfn[v]);
	}
	if(low[u]==dfn[u])
	{
		++scc_cnt;
		for(;;)
		{
			v=s.top();
			s.pop();
			Instack[v]=false;
			sccno[v]=scc_cnt;
			scc[scc_cnt].push_back(v);
			if(v==u) break;
		}
	}
}
void find(int l,int r)
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(Instack,false,sizeof(Instack));
	memset(sccno,0,sizeof(sccno));
	dfs_clock=scc_cnt=0;
	for(int i=l;i<=r;i++)
	if(!dfn[i])
	tarjan(i,-1);
}
void slove()
{
	for(int i=1;i<=n;i++)
	{
		int u=sccno[i];
		int sum=0;
		for(int j=0;j<scc[u].size();j++)
		{
			if(scc[u][j]>n&&map[i][scc[u][j]-n])
			rec[sum++]=scc[u][j]-n;
		}
		sort(rec,rec+sum);
		printf("%d",sum);
		for(int j=0;j<sum;j++)
		printf(" %d",rec[j]);
		printf("\n");
	}
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		init();
		getmap();
		find(1,2*n);
		slove();
	}
	return 0;
}