POJ1904 King's Quest

POJ1904 King's Quest

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 4537		Accepted: 1533
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.

*********************************************************************************

题目大意：一个国王有n个王子，同时有n个女孩。每个王子都有自己喜欢的若干个女孩，现给定一个合法的完备匹配（也就是一个王子娶其中一个自己喜欢女孩），求每个王子可以选择哪些女孩可以让剩下的每个王子依旧能够选择到自己喜欢的一个女孩。

解题思路：参考：http://www.cnblogs.com/zxndgv/archive/2011/08/06/2129333.html

这篇文章写得挺好，一开始我也是用了N次找增广路，直接把增广路算出来的方法，无庸置疑的TLE。然后实在无法，参考了这个博客。

他讲的，我就不说了。说说令我思考的问题：为什么图是这么建的。

这个用强连通的方法建图：男的和所有可以他喜欢的女的连一条有向边；女的，则是根据最后的那个匹配来连一条有向边到她所对应的男的。问：为什么这么建？为什么要用有向边？

为什么这么建！当xi的本来就匹配yi的时候，xi有边到yi，yi也有边到xi，直接就是强连通，这是肯定的。然后，我们需要找增广路的时候，必须是从x出发到y，然后到y的匹配的x，一个y只能有一条边到x，造成了不对称性，也就是，可以有好几个x连到同一个y，但每个y连的都有且只有一个x，这种不对称就是有向图。因为利用匈牙利算法的增广路一定是一条不匹配边一条匹配边这样的交替，所以，这样建图是合理的也是必要的。于是，也就不用解释为什么这个题目要用的是有向图的强连通的求法，而不是无向图的求法。

#include <stdio.h>

#include <string.h>

#include <vector>

#include <algorithm>

#define N 4005

using namespace std;



int mat[N];

vector<int>gra[N],ans;

int low[N],dfn[N],gid[N],now,id,mark[N];

vector<int>st;

int n,a,b,c;



void dfs(int s)

{

    low[s]=dfn[s]=++now;

    st.push_back(s);

    mark[s]=1;

    for(int i=0;i<gra[s].size();i++)

    {

        int t=gra[s][i];

        if(!dfn[t])

        {

            dfs(t);

            low[s]=min(low[s],low[t]);

        }

        else

            if(mark[t])

                low[s]=min(dfn[t],low[s]);

    }

    if(dfn[s]==low[s])

    {

        id++;

        while(st.empty()==0)

        {

            int k=st.back();

            gid[k]=id;

            st.pop_back();

            mark[k]=0;

            if(k==s)break;

        }

    }

}



void re(void)

{

    scanf("%d",&n);

    for(int i=1;i<=n*2;i++)

        gra[i].clear();

    for(int i=1;i<=n;i++)

    {

        int m,k;

        scanf("%d",&m);

        for(int j=1;j<=m;j++)

        {

            scanf("%d",&k);

            gra[i].push_back(k+n);

        }

    }

    for(int i=1;i<=n;i++)

    {

        int k;

        scanf("%d",&k);

        mat[i]=k+n;

        mat[k+n]=i;

        gra[k+n].push_back(i);

    }

}



void run(void)

{

    memset(dfn,0,sizeof(dfn));

    memset(low,0,sizeof(low));

    memset(gid,0,sizeof(gid));

    memset(mark,0,sizeof(mark));

    st.clear();

    now=0;id=0;

    for(int i=1;i<=n;i++)

        if(!dfn[i])

            dfs(i);

    for(int i=1;i<=n;i++)

    {

        ans.clear();

        for(int j=0;j<gra[i].size();j++)

            if(gid[i]==gid[gra[i][j]])

                ans.push_back(gra[i][j]-n);

        sort(&ans[0],&ans[0]+ans.size());

        printf("%d",ans.size());

        for(int j=0;j<ans.size();j++)

            printf(" %d",ans[j]);

        puts("");

    }

}



int main()

{

    re();

    run();

    return 0;

}