POJ 3352 Road Construction——边双连通分量

至少增加的边数 =（ 缩点后树总度数为1的结点数 + 1 ）/ 2

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn= 1e4 + 10;
vector<int> g[maxn];
int dfn[maxn], low[maxn], dg[maxn], id;
bool vis[maxn];
int n, m;
void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++id;
    vis[u] = 1;
    for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (v == fa) continue;
        if (!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if (vis[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
}
void find_ebc() {
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(vis, 0, sizeof(vis));
    id = 0;
    for (int i = 1; i <= n; i++) if (!dfn[i]) tarjan(i, -1);
}
int main() {
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; i++) g[i].clear();
        int u, v;
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        find_ebc();
        memset(dg, 0, sizeof(dg));
        for (int u = 1; u <= n; u++) {
            for (int i = 0; i < g[u].size(); i++) {
                int v = g[u][i];
                if (low[u] != low[v]) {
                    dg[low[u]]++;
                }
            }
        }
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            if (dg[i] == 1) {
                cnt++;
            }
        }
        printf("%d\n", (cnt+1)/2);
    }
    return 0;
}