POJ3177 Redundant Paths【tarjan边双联通分量】

LINK

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题目大意

给你一个有重边的无向图图，问你最少连接多少条边可以使得整个图双联通

思路

就是个边双的模板

注意判重边的时候只对父亲节点需要考虑

你就dfs的时候记录一下出现了多少条连向父亲的边就可以了

然后和有向图不一样的是，在这里非树边的更新不用判断点是不是在栈内，因为无向图中没有u可以到达v但是v不能到达u的情况

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//Author: dream_maker
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
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 = 5e3 + 10;
const int M = 1e4 + 10;
struct Edge {
  int u, v, nxt;
} E[M << 1];
int head[N], tot = 0;
int n, m, du[N];
void add(int u, int v) {
  ++tot;
  E[tot].u = u;
  E[tot].v = v;
  E[tot].nxt = head[u];
  head[u] = tot;
}
int dfn[N], low[N], bel[N], ind = 0, cnt_bcc = 0;
stack<int> st;
void tarjan(int u, int fa) {
  dfn[u] = low[u] = ++ind;
  st.push(u);
  bool k = 0;
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].v;
    if (fa == v && !k) {
      k = 1;
      continue;
    }
    if (!dfn[v]) {
      tarjan(v, u);
      low[u] = min(low[u], low[v]);
    } else {
      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() {
  Read(n), Read(m);
  fu(i, 1, m) {
    int u, v;
    Read(u), Read(v);
    add(u, v);
    add(v, u);
  }
  fu(i, 1, n) if (!dfn[i]) tarjan(i, 0);
  for (int i = 1; i <= tot; i += 2) {
    int u = E[i].u, v = E[i].v;
    if (bel[u] != bel[v]) {
      du[bel[u]]++;
      du[bel[v]]++;
    }
  }
  int ans = 0;
  fu(i, 1, cnt_bcc) 
    if (du[i] == 1) ans++;
  ans = (ans + 1) >> 1;
  Write(ans);
  return 0;
}