poj1904 二分图匹配+强连通分量

http://poj.org/problem?id=1904

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.

/**
poj1904 二分图匹配+强连通分量
题目大意：有n个王子和n个公主，每一个王子仅仅能和他喜欢的公主结婚，公主能够和随意王子结婚。如今知道，每一个王子的结婚对象（都有结婚对象），和每一个王子所喜欢的公主，问
           每一个王子能够和那些公主结婚，而且保证每一个王子都得有结婚对象
解题思路：每一个王子向他喜欢的公主连一条有向边，每一个公主向她的结婚对象连一条有向边，求取每一个王子所在的强连通分量，该王子可和他在同一个强连通分量里的公主结婚。满足
           条件。为什么呢？由于每一个王子仅仅能和喜欢的妹子结婚，初始完美匹配中的丈夫和妻子之间有两条方向不同的边能够互达，则同一个强连通分量中的王子数和妹子数一定是相等的，若王子x能够和另外的一个妹子a结婚，妹子a的原配王子y肯定能找到另外一个妹子b结婚，由于假设找不到的话，则x和a必不在同一个强连通分量中。

所以一个王子能够和全部与他同一强连通分量的妹子结婚，而这不会导致同一强连通分量中的其它王子找不到妹子结婚。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stack>
using namespace std;
const int maxn=5005;
int head[maxn],ip;

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

struct note
{
    int v,next;
} edge[400005];

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

int n,cnt[maxn];
int dfn[maxn],low[maxn],belong[maxn],index,cnt_tar,cont[maxn],instack[maxn*2];
stack<int>q;

void tarjan(int u)
{
    dfn[u]=low[u]=++index;
    q.push(u);
    instack[u]=1;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u])
    {
        cnt_tar++;
        int j;
        do
        {
            j=q.top();
            q.pop();
            belong[j]=cnt_tar;
            instack[j]=0;
            cont[cnt_tar]++;
        }
        while(j!=u);
    }
}

void solve()
{
    index=0,cnt_tar=0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(instack,0,sizeof(instack));
    memset(cont,0,sizeof(cont));
    for(int i=1; i<=n; i++)
    {
        if(!dfn[i])
            tarjan(i);
    }
}

int main()
{
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1; i<=n; i++)
        {
            int x,y;
            scanf("%d",&x);
            for(int j=0; j<x; j++)
            {
                scanf("%d",&y);
                addedge(i,y+n);
            }
        }
        for(int i=1; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            addedge(x+n,i);
        }
        solve();
        for(int u=1; u<=n; u++)
        {
            int k=0;
            for(int i=head[u]; i!=-1; i=edge[i].next)
            {
                int v=edge[i].v;
                if(belong[u]==belong[v])
                {
                    cnt[k++]=v-n;
                }
            }
            sort(cnt,cnt+k);
            printf("%d",k);
            for(int i=0; i<k; i++)
            {
                printf(" %d",cnt[i]);
            }
            printf("\n");
        }
    }
    return 0;
}