CUC DFS及其应用 - Virtual Judge

题目大意：有一个n个点的图，求每个点如果被删除后，图中剩余连通块的数量

1<=n<=1e4

思路：也就是求每一个割点出现在几个双联通分量（没有割点的图）中，找双连通分量也是用tarjan，dfn[u]表示u是第几个遍历到的点，low[u]表示从u出发在不经过父子边的情况下能到达的最早的祖先的dfn值，而如果一个点是割点，那么在不经过该点的情况下无法到达该点的祖先也就是low[v]>=dfn[u]，注意特判起始点的情况，它可能是假割点

//#include<__msvc_all_public_headers.hpp>
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int head[N], head2[N];
struct Edge
{
	int v, next;
}e[N*10];
int cnt = 0;
int n,m;
int dfn[N];
void addedge(int u, int v)
{
	e[++cnt].v = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}
struct Node
{
	int id, cn;
}node[N];
int tot = 0;
void init()
{
	tot = 0;
	cnt = 0;
	for (int i = 0; i <= n; i++)
	{
		node[i].id = i;
		node[i].cn = 1;//答案至少是1
		head[i] = -1;
		dfn[i] = 0;
	}
}
int tarjan(int u,int fa)
{
	int lowu = dfn[u] = ++tot;
	int child = 0;
	for (int i = head[u]; ~i; i = e[i].next)
	{
		int v = e[i].v;
		if (!dfn[v])
		{
			child++;
			int lowv = tarjan(v, u);
			lowu = min(lowu, lowv);
			if (lowv >= dfn[u])
			{//该点是割点
				node[u].cn++;
			}
		}
		else if (dfn[v] < dfn[u] && v != fa)
		{
			lowu = min(lowu, dfn[v]);
		}
	}
	if (fa < 0 && child == 1)
	{//假割点
		node[u].cn = 1;
	}
	return lowu;
}
bool cmp(Node a, Node b)
{
	return a.cn == b.cn ? a.id<b.id : a.cn>b.cn;
}
void solve()
{
	init();
	int u, v;
	while(cin>>u>>v)
	{
		if (u == -1 && v == -1)
		{
			break;
		}
		addedge(u, v);
		addedge(v, u);
	}
	tarjan(0, -1);
	sort(node, node + n, cmp);
	for (int i = 0; i < m; i++)
	{
		cout << node[i].id << " " << node[i].cn << endl;
	}
	cout << endl;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	//cin >> t;
	while (cin>>n>>m)
	{
		if (!n&&!m)
		{
			break;
		}
		solve();
	}
	return 0;
}