RevolC FaeLoN UVA - 10972 双联通分量

//这题图不一定联通，开始没看清，然后wa了几发。 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 5010, M = 20010;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
//栈       栈顶
int stk[N], top;
//每个点属于哪个双连通分量
int id[N];
int dcc_cnt;
//是不是桥
bool is_bridge[M];
//每个点度数
int d[N];
void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}
void tarjan(int u, int from)
{
	//初始化时间戳
	dfn[u] = low[u] = ++ timestamp;
	//入栈
	stk[ ++ top] = u;
	//遍历当前点的邻边
	for (int i = h[u]; ~i; i = ne[i])
	{
		int j = e[i];
		if (!dfn[j])
		{
			tarjan(j, i);
			low[u] = min(low[u], low[j]);
			//说明j到不了u
			//说明是桥
			if (dfn[u] < low[j])
				is_bridge[i] = is_bridge[i ^ 1] = true;
		}
		//如果不是反向边
		else if (i != (from ^ 1))
			low[u] = min(low[u], dfn[j]);
	}
	if (dfn[u] == low[u])
	{
		++ dcc_cnt;
		int y;
		do
		{
			y = stk[top -- ];
			id[y] = dcc_cnt;
		}
		while (y != u);
	}
}
int main()
{
	while(cin>>n>>m)
	{
		memset(id,0,sizeof id);
		memset(stk,0,sizeof stk);
		memset(dfn,0,sizeof dfn);
		memset(low,0,sizeof low);
		memset(is_bridge,0,sizeof is_bridge);
		memset(d,0,sizeof d);
		dcc_cnt=timestamp=0;
		idx=top=0;
		memset(h, -1, sizeof h);
		while (m -- )
		{
			int a, b;
			cin >> a >> b;
			add(a, b), add(b, a);
		}
		//防止搜反向边，所以要加-1(也就是从哪个边过来的)
		for(int i=1;i<=n;i++)
			if(!dfn[i])
				tarjan(i,-1);
		//找每个点的度数
		//枚举所有边，找桥
		for (int i = 0; i < idx; i ++ )
			//如果是桥
			if (is_bridge[i])
				//那么就给出边所在的连通分量的编号度数++
				d[id[e[i]]] ++ ;
		//统计有多少度数为1的节点 ，(树中的叶节点)
		if(dcc_cnt==1)
		{
			printf("0\n");
			continue;
		}
		int cnt = 0;
		for (int i = 1; i <= dcc_cnt; i ++ )
			if (d[i] == 1)
				cnt ++ ;
			else if(d[i]==0)
				cnt+=2;
		printf("%d\n", (cnt + 1) / 2);
	}
	return 0;
}