无向图的双连通分量
边双连通分量
#include <bits/stdc++.h>
using namespace std;
const int N = 5000, M = 2e4 + 10;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
bool is_brigd[M];
int id[N], edcc_cnt;
int d[N];
int n, m;
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
void tarjan(int u, int from)
{
dfn[u] = low[u] = ++ timestamp;
stk[++ top] = u;
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]);
if (dfn[u] < low[j])
is_brigd[i] = is_brigd[i ^ 1] = true;
}
else if (i != (from ^ 1))
low[u] = min(low[u], dfn[j]);
}
if (low[u] == dfn[u])
{
++ edcc_cnt;
int y;
do{
y = stk[top --];
id[y] = edcc_cnt;
} while(y != u);
}
}
int main(void)
{
memset(h, -1, sizeof h);
cin >> n >> m;
for (int i = 0; i < m; i ++)
{
int a, b;
cin >> a >> b;
add(a, b), add(b, a);
}
tarjan(1, -1);
for (int i = 0; i < idx; i ++)
if (is_brigd[i])
d[id[e[i]]] ++;
int cnt = 0;
for (int i = 1; i <= edcc_cnt; i ++)
if (d[i] == 1)
cnt ++;
cout << (cnt + 1) / 2 << endl;
return 0;
}