求解无向图中的双连通分量
#include <bits/stdc++.h>
#define io cin.tie(0), cout.tie(0), ios::sync_with_stdio(false)
#define LL long long
#define ULL unsigned long long
#define EPS 1e-8
#define INF 0x7fffffff
#define SUB -INF - 1
using namespace std;
const int N = 1005;
int n, m, low[N], dfn;
vector<int> G[N];
void dfs(int u, int fa)
{
low[u] = ++dfn;
for (int i = 0; i < G[u].size(); i++)
{
int v = G[u][i];
if (v == fa)
continue;
if (!low[v])
dfs(v, u);
low[u] = min(low[u], low[v]);
}
}
int tarjan()
{
int degree[N];
memset(degree, 0, sizeof(degree));
for (int i = 1; i <= n; i++)
for (int j = 0; j < G[i].size(); j++)
if (low[i] != low[G[i][j]])
degree[low[i]]++;
int res = 0;
for (int i = 1; i <= n; i++)
if (degree[i] == 1)
res++;
return res;
}
int main()
{
while (~scanf("%d%d", &n, &m))
{
memset(low, 0, sizeof(low));
for (int i = 0; i <= n; i++)
G[i].clear();
for (int i = 1; i <= m; i++)
{
int a, b;
scanf("%d%d", &a, &b);
G[a].push_back(b);
G[b].push_back(a);
}
dfn = 0;
dfs(1, -1);
int ans = tarjan();
printf("%d\n", (ans + 1) / 2);
}
return 0;
}