hdu 4685 Prince and Princess（匈牙利算法 连通分量）

看了别人的题解。须要用到匈牙利算法的强连通算法

#include<cstdio>
#include<algorithm>
#include<vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int MAXN = 1005;
int n, m;
int mb[MAXN], ma[MAXN];
bool vis[MAXN], gl[MAXN][MAXN];
vector<int> eg[MAXN];
vector<int> mmp[MAXN], res;
int dfs(int a)
{
	for (int i = 0; i< eg[a].size(); ++i)
	{
		int v = eg[a][i];
		if (!vis[v])
		{
			vis[v] = 1;
			if (!mb[v] || dfs(mb[v]))
			{
				mb[v] = a;
				ma[a] = v;
				return 1;
			}
		}
	}
	return 0;
}
int hungary(int a)
{
	int cnt = 0;
	memset(mb, 0, sizeof mb);
	for (int i = 1; i<= a; ++i)
	{
		memset(vis, 0, sizeof vis);
		cnt += dfs(i);
	}
	return cnt;
}
int dfn[MAXN], low[MAXN], zu, belong[MAXN];
int nc;
int stk[MAXN], top, isinstk[MAXN];
void tarjan(int u)
{
	dfn[u] = low[u] = nc++;
	stk[top++] = u;
	isinstk[u] = 1;
	for (int i = 0; i< mmp[u].size(); ++i)
	{
		int v = mmp[u][i];
		if (dfn[v] == -1)
		{
			tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (isinstk[v] && low[u] > dfn[v])
		{
			low[u] = dfn[v];
		}
	}
	if (dfn[u] == low[u])
	{
		int v;
		do
		{
			v = stk[--top];
			isinstk[v] = 0;
			belong[v] = zu;
		}while( v != u);
		zu++;
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif
	int t;
	scanf("%d", &t);
	for (int o = 1; o<= t; ++o)
	{
		printf("Case #%d:\n", o);
		scanf("%d%d", &n, &m);
		for (int i = 1; i<= n; ++i)
		{
			int k, a;
			scanf("%d", &k);
			eg[i].clear();
			while (k--)
			{
				scanf("%d", &a);
				eg[i].push_back(a);
			}
		}
		int cna = hungary(n);
		cna = n+m-cna;
		for (int i=n+1; i<= cna; ++i)
		{
			eg[i].clear();
			for (int j = 1; j<= cna; ++j)
			{
				eg[i].push_back(j);
			}
		}
		for (int i=m+1; i<= cna; ++i)
		{
			for (int j = 1; j<= n; ++j)
			{
				eg[j].push_back(i);
			}
		}
		int nmc = hungary(cna);
		for (int i = 1; i<= cna; ++i) mmp[i].clear();
		for (int i = 1; i<= cna; ++i)
		{
			int a = mb[i];
			for (int j = 0; j< eg[a].size(); ++j)
			{
				int v = eg[a][j];
				if (v == i) continue;
				mmp[i].push_back(v);
			}
		}
		nc = 1;
		memset(dfn, -1, sizeof dfn);
		memset(low, -1, sizeof low);
		top = 0;
		zu = 0;
		for (int i = 1; i<= cna; ++i)
		{
			if (dfn[i] == -1)
				tarjan(i);
		}
		memset(gl, 0, sizeof gl);
		for (int i = 1; i<= cna; ++i)
		{
			for (int j = 0; j< eg[i].size(); ++j)
			{
				gl[i][eg[i][j]] = 1;
			}
		}
		for (int i = 1; i<= n; ++i)
		{
			res.clear();
			for (int j = 1; j<= m; ++j)
			{
				if (gl[i][j] && belong[j] == belong[ma[i]])
					res.push_back(j);
			}
			int sz = res.size();
			printf("%d", sz);
			for (int j = 0; j< sz; ++j)
			{
				printf(" %d", res[j]);
			}
			puts("");
		}
	}
	return 0;
}