边连通分量

冗余路径

\(\href{https://www.acwing.com/solution/content/20697/}{边连通分量}\)

\(本题是等价于加入最少边是整个图变成边连通分量(没有桥)\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 5010, M = 200010;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
int id[N], dcc_cnt;
bool is_bridge[M];
int d[N];

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_bridge[i] = is_bridge[i ^ 1] = true;
        }
        else if (i != (from ^ 1))
            low[u] = min(low[u], dfn[j]);
    }
    if (dfn[u] == low[u]) {
        ++dcc_cnt;
        int y;
        do {
            y = stk[top--];
            id[y] = dcc_cnt;
        } while (y != u);
    }
}

int main() {
    IO;
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    tarjan(1, -1);
    for (int i = 0; i < idx; ++i) 
        if (is_bridge[i])
            d[id[e[i]]]++;
    int cnt = 0;
    for (int i = 1; i <= dcc_cnt; ++i)
        if (d[i] == 1) cnt++;
    cout << (cnt + 1) / 2 << '\n';
    return 0;
}