学步园

题目大意：求要连多少条边能消除无向图的桥

解题思路：求出图的双向连通子图，每个子图缩点，形成的树有多少个叶子节点，添加的边就为叶子节点数加1除以2

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

struct node
{
    int v, next;
};

const int maxn = 1010;
int head[maxn], dfn[maxn], low[maxn], degree[maxn], n, m, num, index, e; 
node edge[maxn];

void addEdge(int u, int v);
void tarjan(int u, int p);

int main()
{
	while(scanf("%d %d", &n, &m) != EOF)
	{
		memset(head, -1, sizeof(head));
		memset(dfn, -1, sizeof(dfn));
		memset(degree, 0, sizeof(degree));
		e = 0; index = 0; num = 0;
		int u, v;
		for(int i = 0; i < m; i++)
		{
			scanf("%d %d", &u, &v);
			u--; v--;
			addEdge(u, v);
		}
		tarjan(0, -1);
		for(int i = 0; i < n; i++)
		{
			for(int k = head[i]; k != -1; k = edge[k].next)
			{
				if(low[i] != low[edge[k].v])
					degree[low[i]]++;
			}
		}
		int cnt = 0;
		for(int i = 0; i < index; i++)
		{
			if(degree[i] == 1)
				cnt++;
		}
		printf("%d\n", (cnt + 1) / 2);
	}
	return 0;
}

void addEdge(int u, int v)
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
    edge[e].v = u;
    edge[e].next = head[v];
    head[v] = e++;
}

void tarjan(int u, int p)
{
	dfn[u] = low[u] = index++;
	for(int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].v;
		if(dfn[v] == -1)
		{
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
		}
		else if(v != p)
			low[u] = min(low[u], dfn[v]);
	}
}