P2860 [USACO06JAN]Redundant Paths G 题解 ratjan边双连通分量

题目链接：https://www.luogu.com.cn/problem/P2860

题目大意：

给定一个无向连通图，求至少加几条边，能使其变成一个边双连通图。

解题思路：

边双连通分量缩点后计算度数为 \(1\) 的节点个数，假设有 \(cnt\) 个，则答案为 \((cnt+1)/2\)。

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5050, maxm = 20020;
struct Edge {
    int u, v, nxt;
} edge[maxm];
int n, m, head[maxn], ecnt;

void init() {
    memset(head, -1, sizeof(int)*(n+1));
    ecnt = 0;
}

void addedge(int u, int v) {
    edge[ecnt] = {u, v, head[u]}; head[u] = ecnt++;
    edge[ecnt] = {v, u, head[v]}; head[v] = ecnt++;
}

int dfn[maxn], low[maxn], id[maxn], ts, dcc;
bool vis[maxm];
stack<int> stk;

void tarjan(int u) {
    dfn[u] = low[u] = ++ts;
    stk.push(u);
    for (int i = head[u]; i != -1; i = edge[i].nxt) {
        if (vis[i]) continue;
        vis[i] = vis[i^1] = true;
        int v = edge[i].v;
        if (!dfn[v])
            tarjan(v), low[u] = min(low[u], low[v]);
        else
            low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        dcc++;
        int v;
        do {
            v = stk.top();
            stk.pop();
            id[v] = dcc;
        } while (u != v);
    }
}

int d[maxn];

int main() {
    scanf("%d%d", &n, &m);
    init();
    for (int i = 0; i < m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(u, v);
    }
    tarjan(1);
    for (int i = 0; i < 2*m; i += 2) {
        int u = edge[i].u, v = edge[i].v;
        if (id[u] != id[v])
            d[id[u]]++, d[id[v]]++;
    }
    int cnt = 0;
    for (int i = 1; i <= dcc; i++) if (d[i] == 1) cnt++;
    printf("%d\n", (cnt + 1) / 2);
    return 0;
}