Author: lllyouo
Date: 20250413
tag: tarjan、边双连通分量
link: https://www.acwing.com/problem/content/366/

问题描述

link

分析

略

参考代码

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

const int N = 1e5 + 10, M = 4e5 + 10;
int h[N], e[M], ne[M], idx;
int hc[N], ec[M], nec[M], idxc;
int dfn[N], low[N], num;
int c[N], dcc;
int p[N], d[N], fa[N];
bool bridge[M];
int n, m, q, ans;

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

void addc(int a, int b) {
    ec[idxc] = b, nec[idxc] = hc[a], hc[a] = idxc++;
}

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]);
    }
}

void dfsc(int x) {
    for (int i = hc[x]; i != -1; i = nec[i]) {
        int y = ec[i];
        if (!d[y]) {
            d[y] = d[x] + 1;
            fa[y] = x;
            dfsc(y);
        }
    }
}

int get(int x) {
    if (x == p[x]) return x;
    return p[x] = get(p[x]);
}

int main() {
    int T = 0;
    while (cin >> n >> m) {
        ++T;
        if (n == 0 && m == 0) break;
        memset(h, -1, sizeof h);
        memset(hc, -1, sizeof hc);
        memset(dfn, 0, sizeof dfn);
        memset(bridge, 0, sizeof bridge);
        memset(d, 0, sizeof d);
        memset(c, 0, sizeof c);
        idx = idxc = 0;
        dcc = 0;
        
        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]) continue;
            addc(c[x], c[y]), addc(c[y], c[x]);
        }
        
        ans = dcc - 1;
        d[1] = 1;
        dfsc(1);
        
        for (int i = 1; i <= dcc; i++) p[i] = i;
        
        cin >> q;
        cout << "Case " << T << ":" << endl;
        while (q--) {
            int x, y; cin >> x >> y;
            x = c[x], y = c[y];
            if (x != y) {
                x = get(x), y = get(y);
                while (x != y) {
                    if (d[x] < d[y]) swap(x, y);
                    if (x == 1) break;
                    p[x] = get(fa[x]);
                    ans--;
                    x = get(x);
                }
            }
            cout << ans << endl;
        }
        cout << endl;
    }
    

    return 0;
}