poj1904(强连通分量)

King's Quest

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 8460		Accepted: 3087
Case Time Limit: 2000MS

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.

题意：输入n，表示n个王子，接下来n行，每行首先输入nn，表示王子i喜欢的女子有nn个，然后输入这nn个女子的编号。最后一行输入媒婆已经安排好的配对（都是各自王子喜欢的姑娘）。国王要求在每个儿子都能娶到自己喜欢的姑娘的前提下，每个儿子能够娶的姑娘编号。 

思路：由于要输出每个王子能配对的姑娘，所以用二分匹配的思路似乎做不了这题，所以建图的时候让王子先给喜欢的姑娘连边，再在媒婆给出的配对里面让姑娘分别对王子连边，然后缩点，缩完点之后看王子和哪些姑娘在同一个强连通分量里边，如果在一个强连通分量里面直接输出就好。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int N=4010;
struct data
{
    int to,next;
} tu[N*100];
int head[N];
int ip;
int dfn[N], low[N];///dfn[]表示深搜的步数，low[u]表示u或u的子树能够追溯到的最早的栈中节点的次序号
int sccno[N];///缩点数组，表示某个点对应的缩点值
int step;
int scc_cnt;///强连通分量个数
void init()
{
    ip=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
    tu[ip].to=v,tu[ip].next=head[u],head[u]=ip++;
}
vector<int> scc[N];///得出来的缩点，scc[i]里面存i这个缩点具体缩了哪些点
stack<int> S;
void dfs(int u)
{
    dfn[u] = low[u] = ++step;
    S.push(u);
    for (int i = head[u]; i !=-1; i=tu[i].next)
    {
        int v = tu[i].to;
        if (!dfn[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if (!sccno[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u])
    {
        scc_cnt += 1;
        scc[scc_cnt].clear();
        while(1)
        {
            int x = S.top();
            S.pop();
            if (sccno[x] != scc_cnt) scc[scc_cnt].push_back(x);
            sccno[x] = scc_cnt;
            if (x == u) break;
        }
    }
}
void tarjan(int n)
{
    memset(sccno, 0, sizeof(sccno));
    memset(dfn, 0, sizeof(dfn));
    step = scc_cnt = 0;
    for (int i = 1; i <=n; i++)
        if (!dfn[i]) dfs(i);
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1; i<=n; i++)
        {
            int nn;
            scanf("%d",&nn);
            while(nn--)
            {
                int t;
                scanf("%d",&t);
                add(i,t+n);
            }
        }
        for(int i=1; i<=n; i++)
        {
            int t;
            scanf("%d",&t);
            add(t+n,i);
        }
        tarjan(n*2);
        for(int i=1; i<=n; i++)
        {
            int s[N*10],ans=0;
            for(int j=head[i]; j!=-1; j=tu[j].next)
            {
                int to=tu[j].to;
                if(sccno[i]==sccno[to])
                    s[ans++]=to-n;
            }
            sort(s,s+ans);
            printf("%d",ans);
            for(int i=0; i<ans; i++)
                printf(" %d",s[i]);
            puts("");
        }
    }
    return 0;
}