经典傻逼套路
就是把所有边双缩点之后叶子节点的个数
//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e5 + 10;
struct Edge {
int v, nxt;
} E[N << 1];
int head[N], tot = 0, cnt_bcc = 0;
int dfn[N], low[N], bel[N], ind = 0;
stack<int> st;
int n, m, fro[N], to[N], du[N];
void add(int u, int v) {
E[++tot] = (Edge) {v, head[u]};
head[u] = tot;
}
void tarjan(int u, int fa) {
dfn[u] = low[u] = ++ind;
int k = 0;
st.push(u);
for (int i = head[u]; i; i = E[i].nxt) {
int v = E[i].v;
if (v == fa && !k) {
k = 1; continue;
}
if (!dfn[v]) tarjan(v, u), low[u] = min(low[u], low[v]);
else if (dfn[v] < dfn[u]) low[u] = min(low[u], dfn[v]);
}
if (low[u] == dfn[u]) {
int now;
++cnt_bcc;
do {
now = st.top(); st.pop();
bel[now] = cnt_bcc;
} while (now != u);
}
}
int main() {
#ifdef dream_maker
freopen("input.txt", "r", stdin);
#endif
Read(n), Read(m);
fu(i, 1, m) {
Read(fro[i]), Read(to[i]);
add(fro[i], to[i]);
add(to[i], fro[i]);
}
fu(i, 1, n) if (!dfn[i]) tarjan(i, 0);
fu(i, 1, m) {
if (bel[fro[i]] != bel[to[i]]) {
++du[bel[fro[i]]];
++du[bel[to[i]]];
}
}
int ans = 0;
fu(i, 1, cnt_bcc)
if (du[i] == 1) ++ans;
Write((ans + 1) >> 1);
return 0;
}