BZOJ1718: [Usaco2006 Jan] Redundant Paths 分离的路径【边双模板】【傻逼题】

LINK

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经典傻逼套路

就是把所有边双缩点之后叶子节点的个数

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//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;
}