Author: lllyouo
Date: 20250423
tag: tarjan、边双连通分量
link: https://www.luogu.com.cn/problem/P2860

问题描述

link

分析

求边双、缩点之后，统计叶子节点个数，除于 $2$ 之后向上取整。

参考代码

#include <bits/stdc++.h>
using namespace std;

const int N = 5010, M = 10010;
int h[N], e[M * 2], ne[M * 2], idx;
int dfn[N], low[N], num;
int c[N], dcc;
int g[N][N], deg[N];
bool bridge[M * 2];
int n, m, ans;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void tarjan(int x, int in_edge) {
    dfn[x] = low[x] = ++num;
    for (int i = h[x]; i != -1; i = ne[i]) {
        int y = e[i];
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (low[y] > dfn[x]) bridge[i] = bridge[i ^ 1] = true;
        } else if (i != (in_edge ^ 1)) {
            low[x] = min(low[x], dfn[y]);
        }
    }
}

void dfs(int x) {
    c[x] = dcc;
    for (int i = h[x]; i != -1; i = ne[i]) {
        if (!bridge[i] && !c[e[i]]) dfs(e[i]);
    }
}

int main() {
    memset(h, -1, sizeof h);
	cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        add(a, b), add(b, a);
    }

    tarjan(1, 0);

    for (int i = 1; i <= n; i++) {
        if (!c[i]) {
            ++dcc;
            dfs(i);
        }
    }

    for (int i = 1; i < idx; i++) {
        int x = e[i ^ 1], y = e[i];
        if (c[x] != c[y]) {
            g[c[x]][c[y]] = g[c[y]][c[x]] = 1;
        }
    }

    for(int i = 1; i <= dcc; i++) {
        for(int j = 1; j <= dcc; j++) {
            if(i != j && g[i][j]) deg[i]++;
        }
    }

    int ans = 0;
    for(int i = 1; i <= dcc; i++) {
        if(deg[i] == 1) ans++;
    }
	cout << ceil(ans / 2.0) << endl;

    return 0;
}