hdu4685---Prince and Princess

Prince and Princess

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1165    Accepted Submission(s): 325

Problem Description

There are n princes and m princesses. Princess can marry any prince. But prince can only marry the princess they DO love.
For all princes,give all the princesses that they love. So, there is a maximum number of pairs of prince and princess that can marry.
Now for each prince, your task is to output all the princesses he can marry. Of course if a prince wants to marry one of those princesses,the maximum number of marriage pairs of the rest princes and princesses cannot change.

 

Input

The first line of the input contains an integer T(T<=25) which means the number of test cases.
For each test case, the first line contains two integers n and m (1<=n,m<=500), means the number of prince and princess.
Then n lines for each prince contain the list of the princess he loves. Each line starts with a integer k _i(0<=k _i<=m), and then k _i different integers, ranging from 1 to m denoting the princesses.

 

Output

For each test case, first output "Case #x:" in a line, where x indicates the case number between 1 and T.
Then output n lines. For each prince, first print li, the number of different princess he can marry so that the rest princes and princesses can still get the maximum marriage number.
After that print li different integers denoting those princesses,in ascending order.

 

Sample Input

   
   
   
   

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

 

Sample Output

   
   
   
   

    
    
    
    Case #1:
2 1 2
2 1 2
1 3
1 4
Case #2:
2 1 2
   
   
   
   

 

Source

2013 Multi-University Training Contest 8

 

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Statistic | Submit | Discuss | Note

首先做一次最大匹配，设为cnt，那么对于左边，有n-cnt个王子没有匹配，对于右边，有m-cnt个公主没有匹配，所以我们在左边加上m-cnt个虚拟点，这些点喜欢所有公主，右边加上n-cnt个虚拟点，这些点被所有王子喜欢，这样左右两边都是n+m-cnt个点，在求一次最大匹配，这一定是一个完备匹配，对于每一个王子，用他目前匹配的公主，向所有他喜欢的公主连一条有向边，这表示单方面可以替换，所以再对得到的新图求强连通，处在一个强连通分量的公主可以相互替换

/*************************************************************************
    > File Name: hdu4685.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年01月21日 星期三 15时10分15秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1100;

int head[N];
int DFN[N];
int low[N];
int block[N];
int match[N];
int match2[N];
bool used[N];
bool instack[N];
bool g[N][N];
stack <int> st;
vector <int> vis;
int n, m, tot, sccnum, ord, num_v, num_u;

struct node
{
	int next;
	int to;
}edge[N * N * 5];

void init ()
{
	vis.clear();
	memset (g, 0, sizeof(g));
	memset (match, -1, sizeof(match));
	memset (match2, -1, sizeof(match2));
	memset (used, 0, sizeof(used));
	memset (instack, 0, sizeof(instack));
	memset (DFN, -1, sizeof(DFN));
	memset (low, 0, sizeof(low));
	memset (head, -1, sizeof(head));
	tot = sccnum = ord = 0;
}

void addedge (int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool dfs (int u)
{
	for (int i = 1; i <= num_v; ++i)
	{
		if (!g[u][i])
		{
			continue;
		}
		int v = i;
		if (!used[v])
		{
			used[v] = 1;
			if (match[v] == -1 || dfs (match[v]))
			{
				match[v] = u;
				return 1;
			}
		}
	}
	return 0;
}

int hungry ()
{
	int cnt = 0;
	for (int i = 1; i <= num_u; ++i)
	{
		memset (used, 0, sizeof(used));
		if (dfs(i))
		{
			++cnt;
		}
	}
	return cnt;
}

void tarjan (int u)
{
	DFN[u] = low[u] = ++ord;
	instack[u] = 1;
	st.push(u);
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (DFN[v] == -1)
		{
			tarjan (v);
			low[u] = min (low[v], low[u]);
		}
		else if (instack[v])
		{
			low[u] = min (low[u], DFN[v]);
		}
	}
	int v;
	if (low[u] == DFN[u])
	{
		++sccnum;
		do
		{
			v = st.top();
			st.pop();
			instack[v] = 0;
			block[v] = sccnum;
		}while (u != v);
	}
}

void solve ()
{
	num_u = n;
	num_v = m;
	int cnt = hungry();
	num_u = n + m - cnt;
	num_v = num_u;
	for (int i = n + 1; i <= num_u; ++i)
	{
		for (int j = 1; j <= num_v; ++j)
		{
			g[i][j] = 1;
		}
	}
	for (int i = 1; i <= num_u; ++i)
	{
		for (int j = m + 1; j <= num_v; ++j)
		{
			g[i][j] = 1;
		}
	}
//	printf("%d\n", cnt);
	memset (match, -1, sizeof(match));
	cnt = hungry();
//	printf("%d\n", cnt);
	for (int i = 1; i <= num_v; ++i)
	{
		if (match[i] != -1)
		{
			match2[match[i]] = i;
		}
	}
	for (int i = 1; i <= num_u; ++i)
	{
		for (int j = 1; j <= num_v; ++j)
		{
			if (g[i][j] && match2[i] != j)
			{
				addedge (match2[i], j);
			}
		}
	}
	for (int i = 1; i <= num_v; ++i)
	{
		if (DFN[i] == -1)
		{
			tarjan (i);
		}
	}
	for (int i = 1; i <= n; ++i)
	{
		vis.clear();
		for (int j = 1; j <= m; ++j)
		{
			if (g[i][j] && block[j] == block[match2[i]])
			{
				vis.push_back (j);
			}
		}
		int tmp = vis.size();
		printf("%d", tmp);
		for (int j = 0; j < tmp; ++j)
		{
			printf(" %d", vis[j]);
		}
		printf("\n");
	}
}

int main ()
{
	int t;
	int u, v, num;
	int icase = 1;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &n, &m);
		init();
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &num);
			while (num--)
			{
				scanf("%d", &v);
				g[i][v] = 1;
			}
		}
		printf("Case #%d:\n", icase++);
		solve();
	}
	return 0;
}